
Book_Jj_£j5' 



/ 



ELEMENTS 



i 



OF 



GEOMETRY AND TRIGONOMETRY, 



FROM THE WORKS OP 



A. M. LEGENDRE. 



ADAPTED TO THE COURSE OF MATHEMATICAL INSTRUCTION IN 

THE UNITED STATES, 



BY CHAELES DAYIES, LL.D., 

AUTHOR OP ARITHMETIC, ALGEBRA, PRACTICAL MATHEMATICS FOR PRACTICAL MEN, 

ELEMENTS OP DESCRIPTIVE AND OP ANALYTICAL GEOMETRY, ELEMENTS 

OP DIFFERENTIAL AND INTEGRAL CALCULUS, AND SHADES, 

SHADOWS, AND PERSPECTIVE. 



A. S. BARNES & COMPANY, 
NEW YORK AND CHICAGO. 



18G9. 

1899 

v r Or ,.,»ou\viv> A 







3)m^ v ^mm M M$$®$m$^ 



THE OLDEST AND MOST THOROUGH OF ALL MATHEMATICAL SEEIES 
O o O Q Q LATELY RETISED. 



4 



o Vi n o 



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PREFACE. 



Of the various Treatises on Elementary Geometry 
which have appeared during the present century, that 
of M. Legexdee stands preeminent. Its peculiar merits 
have won for it not only a European reputation, but 
have also caused it to be selected as the basis of many 
of the best works on the subject that have been pub- 
lished in this country. 

In the original Treatise of Legendre, the propositions 
are not enunciated in general terms, but by means of 
the diagrams employed in their demonstration. This 
departure from the method of Euclid is much to be 
regretted. The propositions of Geometry are general 
truths, and ought to be stated in general terms, without 
reference to particular diagrams. In the following work, 
each proposition is first enunciated in general terms, and 
afterwards, with reference to a particular figure, that 
figure being taken to represent any one of the class to 
which it belongs. By this arrangement, the difficulty 
experienced by beginners in comprehending abstract truths, 
is lessened, without in any manner impairing the gener- 
ality of the truths evolved. 

The term solid, used not only by Legendre, but by 
many other authors, to denote a limited portion of space, 
6eems calculated to introduce the foreign idea of matter 



iv PREFACE. 

into a science, which deals only with the abstract pro- 
perties and relations of figured space. The term volume, 
has been introduced in its place, under the belief that 
it corresponds more exactly to the idea intended. Many 
other departures have been made from the original text, 
the value and utility of which have been made manifest 
in the practical tests to which the work has been sub- 
jected. 

In the present Edition, numerous changes have been 
made, both in the Geometry and m the Trigonometry. The 
definitions have been carefully revised — the demonstrations 
have been harmonized, and, in many instances, abbreviated — 
the principal object being to simplify the subject as much as 
possible, without departing from the general plan. These 
changes are due to Professor Peck, of the Department of 
Pure Mathematics and Astronomy in Columbia College. For 
his aid, in giving to the work its present permanent form, I 
tender him my grateful acknowledgements. 

CHARLES DAYJES. 

Columbia College, 
Nbw Yoek, April, 1862. 



CONTENTS. 



GEOMETRY. 

FAGB. 

T 9 

Introduction 

BOOK I. 

Definitions, 13 

Propositions, 

BOOK II. 
Ratios and Proportions, 50 

BOOK III. 

The Circle, and the Measurement of Angles, 69 

Problems relating to the First and Third Books, 82 

BOOK IV. 

Proportions of Figures — Measurement of Areas, 93 

Problems relating to the Fourth Book, 129 

BOOK V. 
Regular Polygons — Measurement of the Circle, 136 

BOOK VI. 
Planes, and Polyedral Angles, 157 

BOOK VH. 
Polyedrona, 178 



vi CONTENTS. 

BOOK VIII. 

PAGE. 

Cylinder, Cone, and Sphere, , 210 

EOOK IX. 
Spherical Geometry, 235 

PLANE TRIGONOMETRY. 

INTRODUCTION. 

Definition of Logarithms, 3 

Rules for Characteristics, 4 

General Principles, 5 

Table of Logarithms, 7 

Manner of Using the Table, 8 

Multiplication by Logarithms, 11 

Division by Logarithms, 12 

Arithmetical Complement, 13 

Raising to Powers by Logarithms, 15 

Extraction of Roots by Logarithms, 16 

PLANE TRIGONOMETRY. 

Plane Trigonometry Defined, 17 

Functions of the Arc, 18-22 

Table of Natural Sines, 22 

Table of Logarithmic Sines, 22 

Use of the Table, 23-27 

Solution of Right-angled Triangles, 27-35 

Solution of Oblique-angled Triangles, 36-4 7 

Problems of Application, 48 

ANALYTICAL TRIGONOMETRY. 

Analytical Trigonometry Defined, • 51 

Definitions and General Principles, 5 1-54 

Rules for Signs of the Functions, 54 



CONTENTS. Y b 

PAGE. 

Limiting value of Circular Functions, 55 

Relations of Circular Functions, 57-59 

Functions of Negative Arcs, 60-G2 

Particular values of Certain Functions, 63 

Formulas of Relation between Functions and Arcs, 64-66 

Functiony of Double and Half Arcs, 67 

Additional Formulas, 68-70 

Method of Computing a Table of Natural Sines, 7] 



SPHERICAL TRIGONOMETRY. 

Spherical Trigonometry Defined, 73 

General Principles, 73 

Formulas for Right-angled Triangles, 74-76 

Napier's Circular Parts, 77 

Solution of Right-angled Spherical Triangles, 80-83 

Quadrantal Triangles, 84 

Formulas for Oblique-angled Triangles, 85-92 

^Solution of Oblique-angled Triangles, 92- 104 



MENSURATION. 



Mensuration Defined, 1Q5 

The Area of a Parallelogram, 106 

The Area of a Triangle, 206 

Formula for the Sine of Half an Angle, 108 

Area of a Trapezoid, 222 

Area of a Quadrilateral, 212 

Area of a Polygon, U3 

Area of a Regular Polygon, U4 

To find the Circumference of a Circle, HG 

To find the Diameter of a Circle, HG 

To find the length of an Arc, 117 

Area of a Circle, 117 

Area of a Sector, 118 

Area of a Segment, 118 

Area of a Circular Ring, 119 



I, 

viii CONTENTS. 

PAGE. 

Area of the Surface of a Prism, 120 

Area of the Surface of a Pyramid, 120 

Area of the Frustum of a Cone, 121 

Area of the Surface of a Sphere, 122 

Area of a Zone, 122 

Area of a Spherical Polygon, 123 

Volume of a Prism, 124 

Volume of a Pyramid, 124 

Volume of the Frustum of a Pyramid, 125 

Volume of a Sphere, 126 

Volume of a Wedge, 127 

Volume of a Prismoid, 128 

Volumes of Kegular Polyedrons, 132 



ELEMENTS 



OF 



GEOMETRY 



INTRODUCTION. 

1. Quantity is anything which can be increased, dimin- 
ished, and measured. 

To measure a thing is to find out how many times it 
contains some other thing of the same kind, taken as a stand- 
ard. The assumed standard is called the unit of measure. 

2. In Geometry, there are four species of quantity, viz. : 
Lines, Surfaces, Volumes, and Angles. These are called, 
Geometrical Magnitudes. 

Since the unit of measure is a quantity of the same kind 
as the thing measured, there are four kinds of units of 
measure, viz. : Units of Length, Units of Surface, Units of 
Volume, and Units of Angular Measure. 

3. Geometry is that branch of Mathematics which treats 
of the properties and relations of the Geometrical Magnitudes. 

4. In Geometry, the quantities considered are generally 
represented by pictorial symbols. The operations to be 
performed upon them, and the relations between them, are 
indicated by signs, as in Analysis. 



10 GEOMETRY. 

The following are the principal signs employed : 

The Sign of Addition, + , called plus : 

Thus, A + B, indicates that B is to be added to A. 

The Sign of Subtraction, — , called minus : 
Thus, A — B, indicates that B is to be subtracted 
from A. 

The Sign of Multiplication, X : 

Thus, ixl indicates that A is to be multiplied 

*>y B. 

The Sign of Division, -f- : 

Thus, A^-JB, or, ^, indicates that A is to be 

divided by B. 

The Exponential Sign : 

Thus, A 3 , indicates that A is to be taken three times 
as a factor, or raised to the third power. 

The Radical Sign, \/~ : 

Thus, JA, \/B, indicate that the square root of A, 
:and the cube root of B, are to be taken. 

When a compound quantity is to be operated upon as a 
^single quantity, its parts are connected by a vinculum or 
Vby a parenthesis : 

Thus, A + B X C, indicates that the sum of A and 
B is to be multiplied by G ; and {A + B) + C, indi- 
cates that the sum of A and B is to be divided by C. 

A number written before a quantity, shows how many 
• hues it is to be taken. 

Thus, Z{A-\-B), indicates that the sum of A and B 
is to be taken three times. 

The Sign of Equality, = : 

Thus, A == B + C, indicates that A is equal to the 

sum of B and C. 



INTRODUCTION. 11 

The expression, A = I> + C, is called an equation. The 
part on the left of the sign of equality, is called the first 
member : that on the right, the second member. 

The Sign of Inequality, < : 

Thus, JA < l/B, indicates that the square root of A 
is less than the cube root of B. The opening of the sign 
is towards the greater quantity. 

The sign, . • . is used as an abbreviation of the word 
hence, or consequently. 

5. The general truths of Geometry are deduced by a 
course of logical reasoning, the premises being definitions and 
principles previously established. The course of reasoning 
employed in establishing any truth or principle, is called a 
demonstration. 

6. A Theorem is a truth requiring demonstration. 

7. An Axiom is a self-evident truth. 

8. A Problem is a question requiring a solution. 

9. A Postulate is a self-evident problem. 

Theorems, Axioms, Problems, and Postulates, are all called 
Propositions. 

10 A Lemma is an auxiliary proposition. 

11. A Corollary is an obvious consequence of one or 
more propositions. 

12. A Scholium is a remark made upon one or more 
propositions, with reference to their connection, their use, 
their extent, or their limitation. 



12 GEOMETRY. 

13. An Hypothesis is a supposition made, either in the 
statement of a proposition, or in the course of a demonstra- 
tion. 

14. Magnitudes are equal to each other, when each con- 
tains the same unit an equal number of times. 

15. Magnitudes are equal in all their parts, when they 
may be so placed as to coincide throughout their whole 
extent. 



ELEMENTS OF GEOMETRY. 



BOOK I. 

ELEMENTARY PRINCIPLES. 
DEFINITIONS. 

1. Geometry is that branch of Mathematics which treats 
of the properties and relations of Geometrical Magnitudes. 

2. A Point is that which has position, but not magni- 
tude. 

3. A Line is that which has length, but neither breadth 
nor thickness. 

Lines are divided into two classes, straight and curved. 

4. A Straight Line is one which does not change its 
direction at any point. 

5. A Curved Line is one which changes its direction at- 
" every point. 

The word line, alone, is used for straight line; and the 
word curve, alone, for curved line. 

0. A line made up of straight lines, not lying in the same 
direction, is called a broken line. 

7. A Surface is that which has length and breadth 
without thickness. 



14 



GEOMETRY. 



Surfaces are divided into two classes, plane and curved 
surfaces. 

8. A Plane is a surface, such, that if any two of its' 
points be joined by a straight line, that line will lie wholly 
in the surface. 

9. A Curved Surface is a surface which is neither a 
plane nor composed of planes. 

10. A Plane Angle is the amount of divergence of two 
lines lying in the same plane. 

Thus, the amount of divergence of the 
lines AB and A (7, is an angle. The 



o 



A- 



B 




lines AB and A G are called sides, and 
their common point A, is called the ver- 
tex. An angle is designated by naming its sides, or some- 
times by simply naming its vertex ; thus, the above is called 
the angle BAC, or simply, the angle A. 

11. "When one straight line meets 
another the two angles which they form 
are called adjacent angles. Thus, the A 
angles ABB and BBC are adjacent. 

12. A Right Angle is formed by one 
straight line meeting another so as to 
make the adjacent angles equal The first 
lino is then said to be perpendicular to the second. 

13. An Oblique Angle is formed by 
one straight lino meeting another so as 
to make the adjacent angles unequal. ' 

Oblique angles are subdivided into two classes, acute 
angles, and obtuse angles. 

14. An Acute Angle is less than a 
right angle 




BOOK I. 15 



15. An Obtuse Angle is greater than 



a right angle. 



16. Two straight lines are parallel, 
when they lie in the same plane and can- 
not meet, how far soever, either way, both 

may be produced. They then have the same direction. 

17. A Plane Figure is a portion of a plane bounded 
by lines, either straight or curved. 

18. A Polygon is a plane figure bounded by straight 

lines. 

The bounding lines are called sides of the polygon. The 
broken line, made up of all the sides of the polygon, is called 
the perimeter of the polygon. The angles formed by the 
sides, are called angles of the polygon. 

19. Polygons are classified according to the number of 
their sides or angles. 

A Polygon of three sides is called a triangle ; one of 
four sides, a quadrilateral ; one of five sides, a pentagon ; 
one of six sides, a hexagon ; one of seven sides, a hepta- 
gon ; one of eight sides, an octagon / one of ten sides, a 
decagon ; one of twelve sides, a dodecagon^ &c. 

20. An Equilateral Polygon, is one whose sides are 
all equal. 

An Equiangular Polygon, is one whose angles are nil 
equal. 

A Regular Polygon, is one which is both equilateral 
and equiangular. 

21. Two polygons are mutually equilateral, when their sides, 
taken in the same order, are equal, each to each : that is, fol- 
lowing their perimeters in the same direction, the first side 



16 GEOMETRY. 

of the one is equal to the first side of the other, the second 
side of the one, to the second side of the other, and so 



on. 



22. Two polygons are mutually equiangular, when their 
angles, taken in the same order, are equal each to 
each. 

23. A Diagonal of a polygon is a line joining the ver- 
tices of two angles, not consecutive. 

24. A Base tof a polygon is any one of its sides on 
which the polygon is supposed to stand. 

25. Triangles may be classified with reference either to 
their sides, or their angles. 

When classified with reference to their sides, there are 
two classes : scalene and isosceles. 

1st. A Scalene Triangle is one which 
has no two of its sides equal. 

2d. An Isosceles Triangle is one which 
has two of its sides equal. 

When all of the sides are equal, the 
triangle is equilateral. 

When classified with reference to their angles, there are 
are two classes: right-angled and oblique-angled. 

1st. A Right-angled Triangle is one 
that has one right angle. 

The side opposite the right angle, is called the Mjpolhe- 





nuse. 



2d. An Oblique-angled Triangle is 
one whose angles are all oblique. 




BOOK I. 



17 



If one angle of an oblique-angled triangle is obtuse, the 
triangle is said to be obtuse-angled. If all of the angles 
are acute, the triangle is said to be acute-angled. 

26. Quadrilaterals are classified with reference to the rel- 
ative directions of their sides. There are then two classes 
the first class embraces those which have no two sides par 
allel ; the second class embraces those which have two sides 

parallel. 

Quadrilaterals of the first class, are called trapeziums. 

Quadrilaterals of the second class, are divided into two 
species : trapezoids and parallelograms, 



27. A Trapezoid is a quadrilateral 
which has only two of its sides parallel. 



28. A Parallelogram is a quadrilateral which has its 
opposite sides parallel, two and two. 

There are two varieties of parallelograms : rectangles 
and rhomboids. 



1st. A Rectangle is a parallelogram 
whose angles are all right angles. 



A Square is an equilateral rectangle. 



2d. A Rhomboid is a parallelogram 
whose angles are all oblique. 



A Rhombus is an equilateral rhomboid. 




18 GEOMETRY. 

29. Space is indefinite extension. 

30. A Volume is a limited portion of space. 

AXIOMS. 

1. Things which are equal to the same thing, are equal 
to each other. 

2. If equals be added to equals, the sums will be equal. 

3 If equals be subtracted from equals, the remainders 
will be equal. 

4. If equals be added to unequals, the sums will be 
unequal. 

5. If equals be subtracted from unequals, the remainders 
will be unequal. 

6. If equals be multiplied by equals, the products will be 
equal. 

7. If equals be divided by equals, the quotients will be 
equal. 

8. The whole is greater than any of its parts. 

9. The whole is equal to the sum of all its parts. 

10. All right angles are equal. 

11 Only one straight line can be drawn between two 
points. 

12. The shortest distance between any two points is mea- 
sured on the straight line which joins them. 

13. Through the same point, only one line can be drawn 
parallel to a given line. 



BOOK I. 19 

POSTULATES. 

1. A straight line can be drawn between any two points. 

2. A straight line may be prolonged to any length. 

3. If two lines are unequal, the length of the less may 
be laid off on the greater. 

4. A line may be bisected ; that is, divided into two 
equal parts. 

5. An angle may be bisected. 

6. A perpendicular may be drawn to a given line, either 
from a point without s or from a point on the line. 

7. A line may be drawn, making with a given line an 
angle equal to a given angle. 

8. A line may be drawn through a given point, parallel 
to a given line. 

NOTE. 

In making references, the follcnying abbreviations are employed, viz. \ 
A. for Axiom ; B. for Book ; C. for Corollary ; D. for Definition ; I. 
for Introduction ; P. for Proposition ; Prob. for Problem ; Post, for 
Postulate ; and S. for Scholium. In referring to the same Book, the 
number of the Book is not given ; in referring to any other Book, the 
number of the Book is given. 



20 



GEOMETRY. 




PEOPOSITIOK I. THEOEEM. 

If a straight line meet another straight line, the sum of the 
adjacent angles will be equal to two right angles. 

Let BC meet AB at C: 
then will the sum of the angles 
DC A and BGB be equal to 
two right angles. 

Ai C, let CE be drawn per- 
pendicular to AB (Post. 6) ; then, 
by definition (D. 12), the angles 

EGA and EGB will both be right angles, and conse* 
quently, their sum will be equal to two right angles. 

The angle DGA is equal to the sum of the angles 
EGA and EGB (A. 9) ; hence, 

DGA + JDGB = EGA + EGB + BGB ; 
But, EGB + BGB is equal to EGB (A. 9); hence, 

BGA + BGB = EGA + EGB. 

The sum of the angles EGA and EGB, is equal to 
two right angles ; consequently, its equal, that is, the sum 
of the angles BGA and BGB, must also be equal to two 
right angles ; which was to be proved. 

Cor. 1. If one of the angles BGA, BCB, is a right 
angle, the other must also be a right angle. 

Cor. 2. The sum of the an- 
gles BAG, CAB, BAE, EAF, 
formed about a given point on 
the same side of a straight line 
BE, is equal to two right an- 
gles. For, their sum is equal to 




BOOK I. 



21 



the sum of the angles EAB and EAE; which, from the 
proposition just demonstrated, is equal to two right angles. 

* DEFINITIONS. 

If two straight lines intersect each other, they form four 
angles about the point of intersection, which have received 
different names, with respect to each other. 

1°. Adjacent Angles are 
those which lie on the same side 
of one line, and on opposite sides 
of the other ; thus, A CE and 
ECB, or AGE and A CD, are 
adjacent angles. 

2°. Opposite, or Vektical Angles, are those which lie 
on opposite sides of both lines; thus, ACE and BCB, 
or ACB and ECB, are opposite angles. From the pro- 
position just demonstrated, the sum of any two adjacent 
angles is equal to two right angles. 




PROPOSITION II. THEOREM. 

If two straight lines intersect each other, the opposite or 

vertical angles will be equal. 

Let AB and BE intersect 
at C : then will the opposite 
or vertical angles be equal. 

The sum of the adjacent angles 
ACE and ACB, is equal to 
two right angles (P. I.) : the sum 

of the adjacent angles ACE and ECB, is also equal to 
two right angles. But things which arc equal to the same 
thing, are equal to each other (A. 1) ; hence, 




22 



GEOMETRY. 




ACE -\- ACB = ACE + ECB; 

Taking from both the common 
angle ACE (A. 3), there re- 
mains, 

ACB = ECB. 

In like manner, we find, 

A CD + ACE = ACB + BCB ; 
and, taking away the common angle ACB, we have, 

ACE = BCB. 

Hence, the proposition is proved. 

Cor. 1. If one of the angles abont C is a right angle, 
all of the others will be right angles also. For, (P. L, C. 1), 
each of its adjacent angles will 
be a right angle ; and from the 
proposition just demonstrated, its 
opposite angle will also be a right 



D 



A- 



angle. 



C 



-B 



Cor. 2. If one line BE, is E 

perpendicular to another AB, then will the second line AB 
be perpendicular to the first BE. For, the angles BCA 
and BCB are right angles, by definition (D. 12) ; and 
from what has just been proved, the angles ACE and 
BCE are also right angles. Hence, the two lines are 
mutually perpendicular to each other. 

Cor. 3. The sum of all the 
angles A CB, B CD, B CE, ECE, 

EGA, that can be formed about 
a point, is equal to four right 
angles. 





BOOK I. 23 

For, if two lines be drawn through the point, mutually 
perpendicular to each other, the sum of the angles which 
they form will be equal to four right angles, and it will 
also" be equal to the sum of the given angles (A. 9). Hence, 
the sum of the given angles is equal to four right angles. 

PEOPOSITION 131. THEOBEM. 

If two straight lines have two points in common, they will 
coincide throughout their whole extent, and form one and 
the same line. 

Let A and JB be two points 
common to two lines : then will 
the lines coincide throughout. 

Between A and JB they must 
coincide (A. 11). Suppose, now, that they begin to separate 
at some point (7, beyond AJB, the one becoming ACE, 
and the other A CD. If the lines do separate at C, one 
or the other must change direction at this point; but this 
is contradictory to the definition of a straight line (D. 4) : ^ 
hence, the supposition that they separate at any point is 
absurd. They must, therefore, coincide throughout; which 
was to be proved. 

Cor. Two straight lines can intersect in only one point. 

Note.— The method of demonstration employed above, is 
called the reductio ad absurdum. It consists in assuming an 
hypothesis which is the contradictory of the proposition to 
be proved, and then continuing the reasoning until tha 
assumed hypothesis is shown to be false. Its contradictory i 
thus proved to be true. This method of demonstration is 
often used in Geometry. 




24 GEOMETRY. 



PROPOSITION IV. THEOREM. 

If a straight line meet two other straight lines at a com- 
mon point, making the sum of the contiguous angles 
equal to two right angles, the two lines met will form 
one and the same straight line. 

Let DC meet AG and BC 
at C, making the sum of the 
angles DCA and DCB equal 
to two right angles : then will 
CB be the prolongation of A C. 

For, if not, suppose CE to be the prolongation of AC, 
then will the sum of the angles DC A and DCE be 
equal to two right angles (P. I.) : "We shall, consequently, 
have (A. 1), 

DCA + DCB = DCA + DCE ; 

Taking from both the common angle DCA, there re- 
mains, 

DCB = DCE, . 

which is impossible, since a part cannot be equal to the 
whole (A. 8). Hence, CB must be the prolongation of 
AC ; which teas to be proved. 



PROPOSITION V. THEOREM. 

If two triangles have two sides and the included angle of 
the one equal to two sides and the included angle of 
the other, each to each, the triangles will be equal in all 
their parts. 

In the triangles ABC and DEE, let AB be equal 



BOOK I. 



25 





to BE, AG to BF, and the angle A to the angle D : 
then will the triangles he equal in all their parts. 

For, let ABC be 
applied to BEF, in A 

such a manner that the 
ano-le A shall coincide 
with the angle B, 
the side AB taking 
the direction BE, and 

the side AG the direction BF Then, because AB is 
equal to BE, the vertex i? will coincide with the vertex 
E; and because AG "is equal to BF, the vertex (7 will 
coincide with the vertex F ; consequently, the side BG 
will coincide with the side EF (A. 11). The two triangles, 
therefore, coincide throughout, and are consequently equal in 
all their parts (I., D. 14) ; which teas to be proved. 



PROPOSITION VI. THEOREM. 

If two triangles have two angles and the included side of the 
one equal to tico angles and the included side of the other, 
each to each, the triangles will be equal in all their parts. 



In the triangles 
ABC and BEF, let 
the angle B be equal 
to the angle E, the 
angle G to the angle 
F, and the side BG 
to the side EF: then 




C E 




will the triangles be equal in all their parts. 

For, let ABG be applied to BEF in such a manner 
that the angle B shall coincide with the angle E, the sido 



26 GEOMETRY. 

BG taking the direction FF, and the side BA the direc- 
tion ED. Then, because BG is equal to FF, the vertex 
C will coincide with the vertex F; and because the angle 
G is equal to the angle F, the side GA will take the 
direction FD. Now, the vertex A being at the same time 
on the lines FD and FD, it must be at their intersection 
D (P. m., C.) : hence, the triangles coincide throughout, 
and are therefore equal in all their parts (I., D. 14) ; 
which was to be proved. 



PROPOSITION VII. THEOREM. 

The sum of any two sides of a triangle is greater than the 

third side. 

Let AB G be a triangle : then will 
the sum of any two sides, as AB, BG, 
be greater than the third side AG. 

For, the distance from A to G, 
measured on any broken line AB, BG, 

is greater than the distance measured on the straight line 
AG (A. 12) : hence, the sum of AB and BG is greater 
than A C ; which was to be proved. 

Cor. If from both members of the inequality, 

AG <AB + BG, 

wo take away either of the sides AB, BG, as BG, for 

example, there will remain (A. 5), 

AG - BG<AB; 

that is, the difference between any tioo sides of a triangle is 
less than the third side. 

Scholium. In order that any three given lines may re- 




BOOK I. 2 ? 

present the sides of a triangle, the sum of any two must be 
greater than the third, and the difference of any two must 
be less than the third. 

PROPOSITION VHI. THEOREM. 

If from any point within a triangle two straight lines b 
drawn to the extremities of any side, their sum wiU he 
less than that of the two remaining sides of the triangle. 

Let be any point within the triangle BAG, and let 
the lines OB, OG, be drawn to the 
extremities of any side, as BG : 
then will the sum of BO and OG 
be less than the sum of the sides 
BA and AC. 

Prolong one of the lines, as BO, 
till it meets the side AG in I); then, from Prop. VII., we 

shall have, 

OG < OB 4- BG ; 

adding BO to both members of this inequality, recollecting 
that the sum oi BO and OB is equal to BB, we have 

(A. 4), 

v BO -{-OG <BB + BG. 

From the triangle BAB, we have (P. VII.), 

BB < BA + AB ; 

adding BG to both members of this inequality, recollecting 
that the sum of AB and BG is equal to AG, we have, 

BB + BC< BA + AG. 

But it was shown that BO -f OG is less than BB + BG; 
still more, then, is BO + OG less tl:an BA + AC ; which 
was to be proved. 




28 GEOMETRY. 



PROPOSITION IX. THEOREM. 

If tico triangles have two sides of the one equal to two sides of 
the other, each to each, and the included angles unequal, the 
third sides will be unequal; and the greater side xoill belong 
to the triangle which has the greater included angle. 

In the triangles BAG and BEE, let AB be equal to 
BE, AG to BE, and the angle A greater than the an- 
gle B : then will B G be greater than EF. 

Let the line AG be drawn, making the angle GAG 
eqnal to the angle B (Post. 1) ; make AG equal to BE, 
and draw GG. Then will the triangles AGG and BEF 
have two sides and the included angle of the one equal to 
two sides and the included angle of the other, each to each* 
consequently, GG is equal to EF (P. V.). 

ISTow, the point G may be without the triangle ABG, 
it may be on the side BG, or it may be within the trir 
angle ABG. Each case will be considered separately. 

A D 

1°. When G is 

without the triangle 
ABG. 

In the triangles GIG 
and AIB, we have, 
(P. VII.), 

GI + IG > GG, and BI + IA > AB ; 

whence, by addition, recollecting that the sum of BI and 
IG is equal to BG, and the sum of GI and IA, to GA f 
we have, 

AG + BG > AB 4- GG. 




BOOK I. 



29 



Or, since AG = AB, and GG = EF, we have, 
AB + BG > AB + EF. 
TaHng away the common part AB, there remains (A. 5), 

BG > EF. 



2°. When 6? is on BG. 

In this case, it is obvious 
that GG is less than BG '; or, 
since £ (7 = -EjF 7 , we have, 

2? a > EF. 

3°. When £ is within the triangle ABG. 
From Proposition VIII., we have, 




JBA + BG > 6M + 



(9(7 




or, since GA = BA, and GG = EF, 

wo have, 

ILl + 2? (7 > iVl + EF. 

Taking away the common part AB, 

there remains, 

BG > EF. 

Hence, in each case, BG is greater than JEF; which was 
to be proved. 

Conversely : If in two triangles AB G and DJST^ the 
side AB is equal to the side BE, the side AG to 2>J^ 
and BG greater than JEFJ then will the angle BAG be 
greater than the angle EDF. 

For, if not, BAG must either be equal to, or less than, 
EDF. In the former case, BG would be equal to EF 
(P. V.), and in the latter case, BG would be less than 
EF; either of which would be contrary to the hypothesis : 
hence, BAG must be greater than EDF. 



' 





30 GEOMETRY, 



PKOPOSITION X. THEOllEM. 

If two triangles have the three sides of the one equal to the 
three sides of the other, each to each, the triangles will be 
equal in all their parts. 

In the triangles ABG and BEE, let AB be equal to 
BE, AG to BE, and BG to EF : then will the tri- 
angles be equal in all their parts. 

For, since the sides 
AB, AG, are equal to A D 

BE, BF, each to each, 
if the angle A were 
greater than B, it would 
follow, by the last Pro- 
position, that the side 

BG would be greater than EF; and if the angle A were 
less than B, the side BG would be less than EF. But 
BG is equal to EF, by hypothesis ; therefore, the angle A 
can neither be greater nor less than B : hence, it must be 
equal to it. The two triangles have, therefore, two sides and 
the included angle of the one equal to two sides and the inclu- 
ded angle of the other, each to each ; and, consequently, they 
are equal in all their parts (P. V.) ; which was to be proved. 

Scholium. In triaDgles, equal in all their parts, the equal 
sides lie opposite the equal angles; and conversely. 

4 

PKOPOSITION XI. THEOKEM. 

In an isosceles triangle the angles opposite the equal sides are 

equal. 

Let BAG be an isosceles triangle, having the side AB 
equal to the side A G : then will the angle G be equal to 
the angle B. 



BOOK I. 31 

Join the vertex A and the middle point B of the base 
BC. Then, AB is eqnal to AG, by hypothesis, AD 
common, and BB equal to BC, by 
construction : hence, the triangles BAB, A 

and BAG, have the three sides of the 
one equal to those of the other, each to 
each ; therefore, by the last Proposition, B 
the angle B is equal to the angle G; 
which was to be proved. 

Cor. 1. An equilateral triangle^ is equiangular. 

Cor. 2. The angle BAB is equal to BAG, and BBA 

to GBA : hence, the last two are right angles. Conse- 
quently, a line drawn from the vertex of an isosceles triangle 
to the middle of the base, bisects the vertical angle, and is per- 
pendicular to the base. 

PROPOSITION XII. THEOREM. 

If two angles of a triangle are equal, the sides opposite to 
them are also equal, and consequently, the triangle is isos- 
celes. 

In the triangle ABG, let the angle 
AB G be equal to the angle A CB : 
then will AG be equal to AB, and 
consequently, the triangle will be isosceles. 

For, if AB and AG are not equal, 
suppose one of them, as AB, to be the 
greater. On this, take BB equal to AG (Post. 3), and 
draw BG. Then, in the triangles ABG, BBG, we have 
the side BB equal to AG, by construction, the side BO 
common, and the included angle AGB equal to the included 
angle BBC, by hypothesis: hence, the two triangles are equal 




32 GEOMETRY. 

in aU their parts (P. V.). But this is impossible, because a 

part cannot be equal to the whole (A. 8) : hence, the 

hypothesis that AB and AC are unequal, is false. They 

must, therefore, be equal ; which was to be proved. 

Cor. An equiangular triangle is equilateral. 



PROPOSITION XIII. THEOREM. 

In any triangle, the greater side is opposite the greater angle; 
and, conversely, the greater angle is opposite the greater 
side. 

In the triangle ABC, let the angle 
ACB be greater than the angle ABC: 
then will the side AB be greater than 
the side AC. 

For, draw CD, making the ande 
BCD equal to the angle B (Post. 1) : 
then, in the triangle BOB, we have the angles DCB and 
BBC equal: hence, the opposite sides BB and DC are 
equal (P. XII.). In the triangle ACD, we bave (P. VII.), 

AD + DC > AC', 

or, since DC = DB, and AD + DB = AB, we have, 

AB >AC; 

which icas to he proved. 

Conversely : Let AB be greater than A C : then will the 
angle ACB be greater than the angle ABC. 

For, if ACB were less than ABC, the side AB would 
be less than the side A C, from what has just been proved • 
if ACB were equal to ABC, the side AB would be 
equal to AC, by Prop. XII.; but both conclusions are contrary 




BOOK I. 



33 



to the hypothesis : hence, A CB can neither be less than, 
nor equal to, ABC; it must, therefore, be greater; which 
was to be proved. 




PROPOSITION XIY. THEOREM. 

From a given point only one perpendicular can be drawn to 

a given straight line. 

Let A be a given point, and AB 
a perpendicular to BE: then can no 
other perpendicular to BE be drawn 
from A. 

For, suppose a second perpendicular 
AC to be drawn. Prolong AB till 
BE is equal to AB, and draw CE 
Then, the triangles ABC and EBC will have AB equal 
to BE, by construction, CB common, and the included 
angles ABC and EBC equal, because both are right an- 
gles: hence, the angles ACB and ECB are equal (P. V.) 
But A CB is, by a hypothesis, a right angle : hence, 
ECB must also be a right angle, and consequently, the line 
ACE most be a straight line (P. IV.). But this is impos- 
sible (A. 11). The hypothesis that two perpendiculars can 
be drawn is, therefore, absurd ; consequently, only one such 
perpendicular can be drawn ; which teas to be proved. 

If the given point is on the given line, the proposition 
is equally true For, if from A two perpendiculars AB 
and A C could be drawn to BE, 
we should have BAE and CAB 
each equal to a right angle ; and 
consequently, equal to each other ; 
which is absurd (A. 8). 




34 



GEOMETRY. 



PKOPOSITION XV. THEOEEM. 

If from a 'point without a straight line a perpendicular be 
let fall on the line, and oblique Ivies be drawn to differ- 
ent points of it : 

1°. The perpendicidar will be shorter than any oblique line: 

2°. Any two oblique lines that meet the given line at points 
equally distant from the foot of the perpendicular, wiU 

be equal: 
3°. Of two oblique lines that meet the given line at points 
unequally distant from the foot of the perpendicular, the one 
which meets it at the greater distance will be the longer. 

Let i be a given point, BE a 
given straight line, AB a perpendicular 
to BE, and AB, AC, AE oblique 
lines, B C being equal to BE, and BJD 
greater than BC. Then will AB be 
less than any of the oblique lines, AG 
will be equal to AE, and AD greater 

than A G. 

Prolong AB until BE is equal to AB, and draw 

EC, ED. 

' 1°. In the triangles ABG, FBG, we have the side 
AB equal to BE, by construction, the side BG common, 
and the included angles ABG and FBG equal, because both 
are right angles: hence, EG is equal to AG- (P. V.). 
But, AE is shorter than A CE (A. 12) : hence, AB, the 
half of AE, is shorter than AG, the half of ACE', which 
was to be proved. 

2°. In the triangles ABG and ABE, we have the 
side BG equal to BE, by hypothesis, the side AB com 
mon, and the included angles ABG and ABE equal, 




BOOK I. 



35 



because both are right angles: hence, AC is equal to AE; 
which icas to be proved, v 

3°. It may be shown, as in the first case, that AD is 
equal to DF. Then, because the point C lies within the 
triangle ADF, the sum of the lines AD and DF will be 
greater than the sum of the lines A C and CF (P. YILI.) : 
hence, AD, the half of ADF, is greater than AC, the 
half of A CF ; which was to be proved. 

Cor. 1. The perpendicular is the shortest distance from a 
point to a line. 

Cor. 2. From a given point to a given straight line, only- 
two equal straight lines can be drawn ; for, if there could 
be more, there would be at least two equal oblique lines od 
the same side of the perpendicular ; which is impossible. 



PROPOSITION XVI. THEOREM. 

If a perpendicular be drawn to a given straight line at its 

middle point : 
1°. Any point of the perpendicular will be equally distant ' 

from the extremities of the line: 

2°. Any point, without the perpendicular, will be unequally 
distant from the extremities. 

Let AB be a given straight line, C 
its middle point, and EF the perpendicular. 
Then will any point of EF be equally dis- 
tant from A and B ; and any point without 
EF, will be unequally distant from A and B. 

1°. From any point of EF, as D, draw 
the lines DA and DB. Then will DA 
and DB be equal (P. XV.) : hence, D is 
equally distant from A and B ; which was to be proved. 




36 



GEOMETRY. 



2°. From any point without EF, as I, draw IA and 
IB. One of these lines, as IA, will cut EF in some 
point B ; draw .Z>i?. Then, from what 
has just been shown, BA and BB will 
be equal; but IB is less than the sum 
of IB and BB (P. VII.) ; and because 
the sum of IB and BB is equal to the 
sum of IB and BA, or IA, we have 
ZZ? less than IA : hence, J is unequally 
distant from A and B ; w/wc/t was to be 
proved. 

Cor. If a straight line i£F have two of its points E 
and F equally distant from A and .#, it will be perpen- 
dicular to the line AB at its middle point. 




PROPOSITION XVII. THEOREM. 

If two right-angled triangles have the hypothemtse and a side 
of the one equal to the hypothenuse and a side of the 
other, each to each, the triangles will be equal in all their 
parts. 

Let the right-angled tri- 
angles AB G and BEF have 




the hypothenuse AG equal 
to BF, and the side AB 
equal to BE: then will the 
triangles be equal in all their parts. 

If the side BG is equal to EF, the triangles will be 
equal, in accordance with Proposition X. Let us suppose then, 
that BG and EF are unequal, and that BG is the 
longer. On BG lay off BG equal to EF, and draw 
AG. The triangles ABG and BEF have AB equal to 
BE, by hypothesis, BG equal to EF, by construction, and 



BOOK I. 



37 



the angles B and E equal, because both are right angles; 
consequently, AG is equal to DF (P. V.) But, AC ia 
equal to DF, by hypothesis : hence, A G and A C are equal, 
which is impossible (P. XV.). The hypothesis that B.C and 
EF are unequal, is, therefore, absurd : hence, the triangles 
have all their sides equal, each to each, and are, consequently, 
equal in all of their parts ; ichich was to be proved. 



PROPOSITION XYHI. THEOEEM. 

If two straight lines are perpendicular to a third line, they 

will be parallel. 

Let the two lines A C, BD, be perpendicular to A B : 
then will they be parallel. 

For, if they could meet in a 
point 0, there would be two 
perpendiculars OA, OB, drawn 
from the same point to the same 
straight line ; which is impossible (P. XIV.) : hence, the 
lines are parallel ; which icas to be proved. 



D 



~~~~:::::--o 



A 



"> 
L- 



definitions. 

If a straight line EF inter- 
sect two other straight lines AB 
and CD, it is called a secant, 
twith respect to them. The eight 
angles formed about the points of 
intersection have different names, 
with respect to each other. 

1°. lxTEnion angles on the SAirE side, are those that 
lie on the same side of the secant and within the oilier two 
lines. Thus, BGII and GHD are interior angles on the 
same side. 





38 GEOMETRY. 

2°. Exterior angles on the same side, are those that lie 
on the same side of the secant and without the other two 
lines. Thus, EGB and DHF 
are exterior angles on the same 
side. 

3°. Alternate angles, are 
those that lie on opposite sides 
of the secant and within the 
other two lines, but not adja- 
cent. Thus, AGS and GHB 
are alternate angles. 

4°. Alternate exterior angles, are those that lie on 
opposite sides of the secant and without the , other two lines. 
Thus, AGE and FEB are alternate exterior angles. 

5°. Opposite exterior and interior angles, are those 

that lie on the same side of the secant, the one within and 

the other without the other two lines, but not adjacent. Thus, 

EGB and GEB are opposite exterior and interior angles. 

PROPOSITION XIX. THEOREM. 

If two straight lines meet a third line, malting the stem of 
the interior angles on the same side equal to two right 
angles, the two lines vrill be parallel. 

Let the lines KG and HB meet the line BA, making 
the sum of the angles BAG and ABB equal to two right 
angles : then will KG and HB be parallel. 

Through G, the middle point 
of AB, draw GF perpendicular E g 



to KG, and prolong it to E. 
The sum of the angles GBE 
and GBB is equal to two right lv A F 



G 
C 



BOOK I. 39 

angles (P. I.) ; the sum of the angles FAG and GBB is 
equal to two right angles, by hypothesis : hence (A. l), 

GBE + GBB = FAG + GBB. 

Taking from both the common part GBB, we have the 
angle >GBE equal to the angle FAG. Again, the angle. 
BGE and AGF are equal, because they are vertical an 
gles (P. II.) : hence, the triangles GEB and GFA have 
two of their angles and the included side equal, each to each; 
they are, therefore, equal in all then- parts (P. VI.) : hence, 
the angle GEB is equal to the angle GFA. But, GFA 
is a rio-ht ano-le, bv construction ; GEB must, therefore, be 
a right angle : hence, the lines KG and HB are both per- 
pendicular to EF, and are, therefore, parallel (P. XVIEL) ; 
which iocis to be proved. 

Cor. 1. If two lines are cut by a third line, making the 
alternate angles equal to each other, the two lines will be 
parallel. 

Let the angle UGA be equal 
to GIIB. Adding to both, the 
angle IIGB, we have, 

UGA + EGB = GIIB + UGB. 

But the first sum is equal to 

two right angles (P. I.) : hcnco^ 

the second sum is also equal to two right angles ; therefore, 

from what 1ms just been shown, AB and CB arc parallel. 

Cor. 2. If two lines are cut by a third, making the oppo- 
site exterior and interior angles equal, the two lines will be 
parallel. Let the angles EGB and GIIB be equal : Now 
EGB and A Gil are equal, because they are vertical (P. II.); 
and consequently, AGII and GIIB are equal : hence, from 
Cor. 1, AB and CB are parallel. 




40 



GEOMETRY. 




PROPOSITION XX. THEOREM. 

If a straight line intersect two parallel straight lines, the sum 
of the interior angles on the same side will be equal to 
two right angles. 

Let the parallels AB, CB, be cut by the secant line 
EE : then will the sum of HGB and GHB be equal to 
two right angles. 

For, if the sum of HGB 
and GHB is not equal to 
two right angles, let IGL be 
drawn, making the sum of HGB 
and GHB equal to two right 
angles : then IB and CB will 
be parallel (P. XIX.) ; and consequently, we shall have two 
lines GB, GB, drawn through the same point G and par- 
allel to CB, which is impossible (A. 13) : hence, the sum 
of HGB and GHB, is equal to two right angles; which 
was to be proved. 

In like manner, it may be proved that the sum of HGA 
and GHC, is equal to two right angles. 

Cor. 1. If HGB is a right angle, GHB will be a right 
angle also : hence, if a line is perpendicxdar to one of two 
parallels, it is perpendicular to the other also. 

Cor. 2. If a straight line meet two parallels, the alternate 
angles will be equal. 

For, if AB and CB are 
parallel, the sum of BGH and 
GHB is equal to two right 
angles : the sum of BGH and 
HGA is also equal to two right 
angles (P. I.) : hence, these sums 




BOOK I. 41 

are equal. Taking away the common part BGH, there re- 
mains the angle GHD equal to JIG A. In like manner, 
it may be shown that BGH and GHC are equal. 

Cor. 3. If a straight line meet two parallels, the opposite 
exterior and interior angles will be equal. The angles DUG 
and HGA are equal, from what has just been shown. The 
angles HGA and BGE are equal, because they are verti- 
cal : hence, DUG and BGE are equal. In like manner, 
it may be shown that CHG and AGE are equal. 

Scholium. Of the eight angles formed by a line cutting 
two parallel lines obliquely, the four acute angles are equal, 
and so, also, are the four obtuse angles. 



PROPOSITION XXI. THEOREM. . 

If tico straight lines intersect a third line, making the sum 
of the interior angles on the same side less than two right 
angles, the two lines will meet if sufficiently produced. 

Let the two lines CD, II, meet the line EF, making 
the sum of the interior angles IIGL, GHD, less than two 
right angles : then will IL and CD meet if sufficiently pro- 
duced. 

For, if they do not meet, 
they must be parallel (D. 16). 
But, if they were parallel, the 
sum of the interior angles IIGI, 
GHD, would be equal to two 
right angles (P. XX.), which is 
contrary to the hypothesis : hence, 
II, CD, will meet if sufficiently produced ; which was to be 
proved. 




42 



geometry: 



Cor. It is evident that IE and CD, will meet on jthat 
side of EF, on which the sum of the two angles is less 
than two right angles. 



PPwOPOSITION XXII. THEOKEM. 

If two straight lines are parallel to a third line, they are 

parallel to each other. 

Let AB and CD be respectively 
parallel to EF: then will they be par- 
allel to each other. 

For, draw PR perpendicular to 
EF; then will it be perpendicular to 
AB, and also to CD (P. XX., C. 1) : 
hence, AB and CD are perpendicu- 
lar to the same straight line, and consequently, they are par- 
allel to each other (P. XVIII.) ; which was to be proved. 









E 


R 


¥ 


C 


Q 


1) 


A 


P 


B 



PKOPOSITION XXIII. THEOKEM. 
Two parallels aie everywhere equally distant. 

Let AB and CD be parallel : then will they be every- 
where equally distant. 

From any two points of AB, as 
F and E, draw FIT and EG 
perpendicular to CD ; they will also be 
perpendicular to AB (P. XX., C. 1), 
and will measure the distance between 

AB and CD, at the points F and E Draw also FG 
The lines FII and EG are parallel (P. XVIII.) : hence, 
the alternate angles HFC and FGE are equal (P. XX., C. 2). 
The hues AB and CD are parallel, by hypothesis : hence, 




BOOK I. 43 

the alternate angles EFG and FGS are equal. The tri- 
angles FGE and FGH have, therefore, the angle EGF 
equal to GEE, GFH equal to FGE, and the side FG 
common ; they are, therefore, equal in all their parts (P. VI.) : > 
hence, FH is equal to EG ; and consequently, AB and 
CD are everywhere equally distant ; which was to be proved. 



PROPOSITION XXIY. THEOREM. 

If two angles have their sides parallel, and lying either in 
the same, or in opposite directions, they toill be equal 

1°. Let the angles ABC and JDEF have their sides 
parallel, and lying in the same direction : then will they b' 

equal. 

Prolong FE to Z. Then, because W / D 

BE and AL are parallel, the exterior / / 

angle BEF is equal to its opposite in- LA %£ F 

terior angle ABE (P. XX., C. 3) ; and £ 

because BC and BE are parallel, the 
exterior angle ABE is equal to its op- 
posite interior angle ABC : hence, BEF is equal to 
ABC ; which was to be proved. 

2°. Let the angles ABC and GHK 
have their sides parallel, and lying in op- 
posite directions : then will they be equal. 

Prolong Gil to 31. ' Then, because 
KII and B3I are parallel, the exterior 
angle GJIK is equal to its opposite interior angle H31B ; 
and because IIM and BC are parallel, the angle II3IB 
is equal to its alternate angle 3IBC (P. XX., C. 2) : hence, 
GIIK is equal to ABC; which was to be iiroved. 

Cor. The opposite angles of a parallelogram arc equal 





U GEOMETRY. 



PROPOSITION XXV. THEOREM. 

In any triangle, the sum of the three angles is equal to two 

right angles. 

Let CBA be any triangle : then will the sum of the 
angles G, A, and B, be equal to 
two right angles. 

For, prolong CA to B, and draw 
AE parallel to BG. 

Then, since AE and CB are 
parallel, and CD cuts them, the ex 
terior angle DAE is equal to its 

opposite interior angle C (P. XX., C. 3). In like manner, 
since AE and CB are parallel, and AB cuts them, the 
alternate angles AB G and BAE are equal : hence, the 
sum of the three angles of the triangle BAG, is equal to 
the sura of the angles CAB, BAE, EAD ; but this sum 
is equal to two right angles (P. I., C. 2); consequently, the 
sum of the three angles of the triangle, is equal to two 
4ght angles (A. 1) ; which was to be proved. 

Cor. 1. Two angles of a triangle being given, the third 
will be found by subtracting their sum from two right angles. 

Cor. 2. If two angles of one triangle are respectively 
equal to two angles of another, the two triangles are mutually 
equiangular. 

Cor. 3. In any triangle, there can be but one right angle; 
for if there were two, the third angle would be zero. Nor 
can a triangle have more than one obtuse angle. 

Cor. 4. In any right-angled triangle, the sum of the acute 
angles is equal to a right angle. 



BOOK I. 45 

Cor. 5. Since every equilateral triangle is also equiangular 
(P. XI., C. 1), each of its angles will be equal to the third part 
of two right angles ; so that, if the right angle is expressed 
by 1, each angle, of an equilateraf triangle, will be expressed 

by |. 

Cor. 6. In any triangle ABC, the exterior angle BAD 
is equal to the suni of the interior opposite angles B and 
C. For, AE being parallel to BC, the part BAJE is 
equal to the angle B, and the other part DAE, is equal 
to the angle C. 



PROPOSITION XXYI. THEOREM. 

The sum of the interior angles of a polygon is equal to 
two right angles taken as many times as the polygon has 
sides, less two. 

Let A B CDE be any polygon : tnen will the sum of its 
interior angles A, B, C, D, and E, be equal to two right 
angles taken as many times as the polygon has sides, less 
two. 

From the vertex of any angle A, draw 
diagonals AC, AD. The polygon will be 
divided into as many triangles, less two, as 
it has sides, having the point A for a 
common vertex, and for bases, the sides of 
the polygon, except the two which form the 
angle A. It is evident, also, that the sum of the angles of 
these triangles does not differ from the sum of the angles of 
the polygon : hence, the sum of the angles of the polygon is 
equal to two right angles, taken as many times as there are 
triangles ; that is, as many times as the polygon has sides, 
less two j which was to be proved. 




46 



GEOMETRY. 



Cor. 1. The sum of the interior angles of a quadrilateral 
is equal to two right angles taken twice ; that is, to four 
right angles. If the angles of a quadrilateral are equal, each 



will he a right angle. 



Cor. 2. The sum of the interior angles of a pentagon is 
equal to two right angles taken three times ; that is, to six 
right angles : hence, when a pentagon is equiangular, each 
angle is equal to the fifth part of six right angles, or to f 



of one right angle. 



Cor. 3. The sum of the interior angles of a hexagon is 
equal to eight right angles : hence, in the equiangular 
hexagon, each angle is the sixth part of eight right angles, 
or 4- of one ri^ht angle. 



>. 



Cor. 4. In any equiangular polygon, any interior angle is 
equal to twice as many right angles as the figure has sides, 
less four, divided by the number of angles. 



PROPOSITION XXVII. THEOREM. 

The sum of the exterior angles of a polygon is equal to 

four right angles. 

Let the sides of the polygon ABCDE 
be prolonged, in the same order, forming 
the exterior angles a, b, c, d, e ; then will 
the sum of these exterior angles be equal 

to four right angles. 

For, each interior angle, together with 
the corresponding exterior angle, is equal 
to two right angles (P. I.) : hence, the sum of all the inte- 
rior and exterior angles is equal to two right angles taken 




BOOK I. 47 

as many times as the polygon has sides. But the sum of 
the interior angles is equal to two right angles taken as 
many times as the polygon has sides, less two : hence, the 
sum of the exterior angles is equal to two right angles taken 
twice; that is, equal to four right angles; which was to be 
proved, 

PEOPOSITION XXVm. THEOREM. 

In any parallelogram, the opposite sides are equal, each to 

each. 

Let AB CB be a parallelogram : then 
will AB be equal to BG, and AB to 

BC. 

For, draw the diagonal BB. Then, 
because AB and BG are parallel, the 
angle BBA is equal to its alternate 
angle BBC (P. XX., C. 2): and, because AB and BG 
are parallel, the angle BBA is equal to its alternate angle 
BBG. The triangles ABB and GBB, have, therefore, 
the angle BBA equal to GBB, the angle BBA equal 
to BBG, and the included side BB common; consequently, 
they are equal in all of their parts : hence, AB is equal 
to BG, and AB to BG', which was to be proved. 

Cor. 1. A diagonal of a parallelogram divides it into two 
equal triangles. 

Cor. 2. Two parallels included between two other par- 
allels, are equal. 

Cor. 3. If two parallelograms have two sides and the 
included angle of the one, equal to two sides and the included 
angle of the other, each to each, they will be equal. 





48 GEOMETRY. 



, PROPOSITION XXIX. THEOREM. 

If the opposite sides of a quadrilateral are equal, each to 
each, the figure is a parallelogram. 

In the quadrilateral ABCB, let AB 
be equal to DC, and AB to BO: 
then will it be a parallelogram. 

Draw the diagonal DB. Then, the 
triangles ABB and CBB, will have 

the sides of the one equal to the sides of the other, each to 
each; and therefore, the triangles will be equal in 'all of their 
parts : hence, the angle ABB is equal to the angle CBB 
(P. X., S.) ; and consequently, AB is parallel to BO (P. 
XIX., C. 1). The angle BBC is also equal to the angle 
BBA, and consequently, B C is parallel to AB : hence, 
the opposite sides are parallel, two and two ; that is, the 
figure is a parallelogram (D. 28) ; which icas to be proved. 



PROPOSITION XXX. THEOREM. 

If tioo sides of a quadrilateral are equal and parallel, the 

figure is a parallelogram. 

\ _ , 

In the quadrilateral ABCB, let AB 

be equal and parallel to BC: then will 
the figure be a parallelogram. 

Draw the diagonal BB. Then, be- 
cause AB and BC are parallel, the 
angle ABB is equal to its alternate angle CBB. Now, 
the triangles ABB and CBB, have the side BC equal 
to AB, by hypothesis, the side BB common, and the 
included angle ABB equal to BBC, from what has just 




BOOK I. 49 

been shown; hence, the triangles are equal in all their parts 
(P. V.) ; and consequently, the alternate angles ADB and 
DBC are equal. The sides BC and AD are, therefore, 
parallel, and the figure is a parallelogram ; which was to be 
proved. 

Cor. If two points be taken at equal distances from a 
line, and on the same side of it, the line joining them will 
be parallel to the given line. 



PROPOSITION XXXI. THEOREM. 

The diagonals of a parallelogram divide each other into 
equal parts, or mutually bisect each other. 

Let AD CD be a parallelogram, and 
A C, BD, its diagonals : then will AE 
be equal to EC, and BE to ED. 

For, the triangles BEG and AED, 
have the angles EBC and ADE equal 
(P. XX., C. 2), the angles ECB and DAE equal, and the 
included sides B C and AD equal : hence, the triangles 
are equal in all of their parts (P. VI.) ; consequently, AE is 
equal to EC, and BE to ED ; which teas to be proved. 

Scholium. In a rhombus, the sides AB, BC, being 
equal, the triangles AEB, EBC, have the sides of the 
one equal to the corresponding sides of the other ; they are, 
therefore, equal: hence, the angles AEB, BEC, are equal, 
and therefore, the two diagonals bisect each other at right 
angles. 




BOOK II. 

BATIOS AND PROPORTIONS, 
DEFINITIONS. 

1. The Ratio of one quantity to another of the same 
kind, is the quotient obtained by dividing the second by the 
first. Vrhe first quantity is called the Antecedent, and the 
second, the Consequent. 

2. A Proportion is an expression of equality between 
two equal ratios. Thus, 

A ~ ' 

expresses the fact that the ratio of A to B is equal to 
the ratio of G to B. In Geometry, the proportion is 
written thus, 

A : B : : G : £>, 

and read, A is to B, as G is to B. 

3. A Continued Proportion is one in which several 
ratios are successively equal to each other ; as, 

A : B : : G : B : : E : F : : G - : -HJ &c 

4. There are four terms in every proportion. The first 
tnd second form the first couplet, and the third and fourth, 



BOOK II. 51 

the second couplet. The first and fourth terms are called 
extremes; the second and third, means, and the fourth term, 
a fourth proportional to the other three. When the second 
term is equal to the third, it is said to be a mean proportional 
between the extremes. In this case, there are but three 
different quantities in the proportion, and the last is said to 
be a third proportional to the other two. Thus, if we have, 

A : B : : B : C, 

B is a mean proportional between A and C, and C is a 
third proportional to A and B. 

5. Quantities are in proportion by alternation, when ante- 



cedent is compared with antecedent, and consequent with con- 
sequent. 

6. Quantities are in proportion by inversion, when ante- 
cedents are made consequents, and consequents, antecedents. 

7. Quantities are in proportion by composition, when the 
sum of antecedent and consequent is compared with either 
antecedent or consequent. 

8. Quantities are proportional by division, when the differ- 
ence of the antecedent and consequent is compared either with 
antecedent or consequent. 

9. Two varying quantities are reciprocally or inversely 
proportional, when one is increased as many times as the 
other is diminished. In this case, their product is a fixed 
quantity, as xy = m. 

10. Equimultiples of two or more quantities, are the pro- 
ducts obtained by multiplying both by the same quantity. 
Thus, mA and mB, are equimultiples of A and B. 



52 GEOMETRY. 



PROPOSITION I THEOREM. 

If four quantities are in proportion, the product of the 
means will he equal to the product of the extremes. 

Assume the proportion, 

A .' : B j : G : D\ whence, -j = -^ ; 

clearing of fractions, we have, 

BG == AD; 
which was to be proved. 

Cor. If J? is equal to C, there will be but three pro- 
portional quantities ; in this case, the square of the mean is 
equal to the product of the extremes. 



PROPOSITION II. THEOREM. 

If the product of two quantities is equal to the product of 
two other quantities, two of them may he made tlie 
means, and the other two the extremes of a proportion. 

If v/e have, 

AD = BG, 

by changing the members of the equation, we have, 

BG = AZ>; 
dividing both members by AC, we have, 

5. - "& , or A : B : : G : D ; 

A 

which was to be proved. 



BOOK II. 53 

PEOPOSITION m. THEOEEM. 

If four quantities are in proportion, they will be in pro- 
portion by alternation. 

Assume the proportion, 

B JD 

A : B : : C : JD \ whence, -j = -^- • 

(j 
Multiplying both members by y^ t we have, 

J = J ; or, A : C :: B : 2> ; 

which was to be proved. 



PEOPOSITION IV. THEOEEM. 

If one couplet in each of two proportions is the same, the 
other couplets will form a proportion. 

Assume the proportions, 

-^ B B 

A : B : : C i B; whence, -j- = -^ ; 

B G- 
and, A : B : : F : G ; whence, -j- = ^ • 

From Axiom 1, we have, 

-— = -= ; whence, C : D : : F : # ; 

which was to be proved. 

Cor. If the antecedents, in two proportions, are the same 
the consequents will be proportional. For, the antecedents 
of the second couplets may be made the consequents of the 
first, by alternation (P. III.). 



54 GEOMETRY. 



x PROPOSITION V. THEOREM. 

If four quantities are in proportion, they will be in pro- 
portion by inversion. 

Assume the proportion, 

t* b 

A : B : : C : B ; whence, -^ = -^ • 
If we take the reciprocals of both members (A. 7), we have, 
ti=~; whence, B : A : : B : C ; 

which was to be proved. 



PROPOSITION VI. THEOREM. 

If four quantities are in proportion, they will be in pro- 
portion by composition or division. 

Assume the proportion, 

B B 

A : B : : C : B\ whence, -j = ~q ' 

If we add 1 to both members, and subtract 1 from both 
members, we shall have, 

B B , B ' B 

£ + 1 = ^ + 1 ; and, x - 1 = jj - 1 ; 

whence, by reducing to a common denominator, we have, 

V±A=^C, and, 5^4 =££-£; whence, 
A 5 A 

A : B+A : : C : JP+ (7, and, J. : J?-4 : * C : B-G 

which was to be proved. 



BOOK II. 55 



PROPOSITION VH. THEOREM. 

Equimultiples of two quantities are proportional to the quan>* 

titles themselves. 

Let A and B be any two quantities ; then -j will 

denote their ratio. 

If we multiply both terms of this fraction by m, its 
value will not be changed ; and we shall have, 

m — r = -r ; whence, mA : mB : : A : B ; 
mA A 

which was to be proved. 



PROPOSITION VTH. THEOREM. 

If four quantities are in proportion, any equimultiples of 
the first couplet will he proportional to any equimultiples 
of the second couplet. 

Assume the proportion, 

A : B : : C : D ; whence, -=- = ^=- • 

' AG 

If we multiply both terms of the first member by m, and 
both terms of the second member by n, we shall have, 

— 7 = — ~ ; whence, mA : mB : : nC : nB ; 
mA nC 

which was to be proved. 



56 GEOMETRY. 



PROPOSITION IX. THEOEEM. 

If two quantities be increased or diminished by like parte 
of each) the results will he proportional to the quantities 
themselves. 

We have, Prop. VII., 

A : B : : mA : mB. 

If we make m — 1 ± — , in which — is any fraction, 

9 2 

we shall have, 

A : B : : A ±^-A : B ±' P -B ; 

9 9 

which was to be proved. 



PROPOSITION X. THEOREM. 

If both terms of the first couplet of a proportion be in- 
creased or diminished by like parts of each ; and if both 
terms of the second couplet he increased or diminished by 
any other like parts of each, the results will be in pro- 
portion. 

Since we have, Prop. VIII., 

m A : mB : : nC : nD ; 

if we make m — 1 ± — , and, n = 1 ± — , we shall 

q q 

have, 

A±2-A : B± P -B :: C ±^C : D ± ^D ; 
q 9 9 9 

which was to be proved. 



BOOK II. 



57 



PEOPOSITION XI. THEOEEM. 

In any continued proportion, the sum of the antecedents is 
to the sum of the consequents, as any antecedent to its 
corresponding consequent. 

From the definition of a continued proportion (D. 3), 



A : B : : C : D : : E : F : : G 
hence, 

7? 7? 

whence, BA = AB 



IT, &o, 



B B 

A ~ A> 



B D 

a ~ a ; 

B _F 

A ~ E ' 

B _2Z\ 

A - G ' 



whence, B C = AD 



whence, BE = AF 



whence, BG = AH 



&c. 



Adding and factoring, we have, 



B{A + C + E+ G + &c) = A{B +D + F+ H + &c.) : 

hence, from Proposition II., 

A + C + E+G + &C. : B + D + F+ ZT+ &c •: A : B ; 



which was to be proved. 



58 GEOMETRY. 



PEOPOSITION XH. THEOKEM. 

If two proportions be multiplied together, term by term, tte 
the products will be proportional. 

Assume the two proportions, 

B B 

A : B : : C : J> ; whence, -j = -^ ; 

F JB" 

and, E \ F \\ G \ H\ whence, -^ = ^ • 

Multiplying the equations, member by member, we have, 

BF_BH h ^ : BF :: CG : DM* 

AE ~ CG> 

which was to be proved. 

Cor. 1. If the corresponding terms of two proportions 
are equal, each term of the resulting proportion will be the 
square of the corresponding term in either of the given pro- 
portions : hence, If four quantities are proportional, their 
squares will be proportional. 

Cor. 2. If the principle of the proposition be extended 
to three or more proportions, and the corresponding terms 
of each be supposed equal, it will follow that, like powers 
qf proportional quantities are proportionals. 




BOOK III. 

THE CIRCLE AND THE MEASUREMENT OF ANGLES 

DEFINITIONS. 

1. A Circle is a plane figure, 
bounded by a curved line, every point 
of which is equally distant from a point 
within, called the centre. 

The bounding line is called the cir- 
cumference. 

2. A Radius is a straight line drawn from the centre 
to any point of the circumference. 

3. A Diameter is a straight line drawn through the 
centre and terminating in the circumference. 

All radii of the same circle are equal. All diameters 
are also equal, and each is double the radius. 

4. An Arc is any part of a circumference. 

5. A Chord is a straight line joining the extremities of 

an arc. 

Any chord belongs to two arcs : the smaller one is meant, 
unless the contrary is expressed. 

G. A Segment is a part of a circle included between an 
arc and its chord. 

7. A Sector is a part of a circle included within an 
an arc and the radii drawn to its extremities. 



60 



GEOMETRY. 



8. An Inscribed Angle is an angle 
whose vertex is in the circumference, and 
whose sides are chords. 






9. An Inscribed Polygon is a poly- 
gon whose vertices are in the circumfer- 
ence, and whose sides are chords. 

10. A Secant is a straight line which 
cuts the circumference in two points. 

11. A Tangent is a straight line 
which touches the circumference in one 
point. This point is called, the point of 
contact, or, the point of tangency. 

12. Two circles are tangent to 
each other, when they touch each 
other in one point. This point is 
called, the point of contact, or the 
point of tangency. 

13. A Polygon is circumscribed about 
a circle, when all of its sides are tangent 
to the circumference. 

14. A Circle is inscribed in a polygon, 
when its circumference touches all of the 
tides of the polygon. 

POSTULATE. 

A circumference can be described from any point as a 
centre, and with any radius. 





BOOK III. 



61 



PROPOSITION I. THEOREM. 



Any diameter divides the circle, and also its circumference, 

into tico equal parts. 

Let AEBF be a circle, and AB 
any diameter : then will it divide the 
circle and its circumference into two 
equal parts. 

For, let AFB be applied to AUB, 
the diameter AB remaining common ; E 

then will they coincide; otherwise there would be some points 
in either one or the other of the curves unequally distant 
from the centre ; which is impossible (D. 1) : hence, AB 
divides the circle, and also its circumference, into two equal 
parts ; which was to he proved. 




PROPOSITION II. THEOREM. 
A diameter is greater than any other chord. 

Let AD be a chord, and AB a diameter through one 
extremitv, as A : then will AB be greater than AB. 

Draw the radius CD. In the tri- 
angle A CD, we have AD less than 
the sum of AC and CD (B. I., P. 
VII.) . But this sum is equal to 
AB (D. 3) : hence, AB is greater 
than AD ; lohich was to be proved. 




62 



GEOMETRY. 



PROPOSITION HI. THEOREM. 

A straight line cannot meet a circumference in more than 

two points. 

Let AEBF be a circumference, and 
AB a straight line : then AB cannot 
meet the circumference in more than two 
points. 

For, suppose that they could meet in 
three points. We should then have three 
equal straight lines drawn from the same point to the same 
straight line ; which is impossible (B. I., P. XV., C. 2) : 
hence, AB cannot meet the circumference in more than 
two points ; which was to be proved. 




PROPOSITION IV. THEOREM. 

In equal circles, equal arcs are subtended by equal chords ; 
and conversely, equal chords subtend equal arcs. 

1°. In the equal cir- 
cles ABB and EGF, 
let the arcs AMD and 
ENG be equal : then 
will the chords AB and 
EG be equal. 

Draw the diameters AB and EF. If the semi-circle 
ABB be applied to the semi-circle EGF, it will coincide 
with it, and the semi-circumference ABB will coincide with 
the semi-circumference EGF. But the part AMB is equal 
to the part ENG, by hypothesis : hence, the point B will 
fall on G; therefore, the chord AB will coincide with 




BOOK III. 



63 



EG (A. 11), and is, therefore, equal to it; which was to 
be proved. 

2°. Let the chords AD and EG be equal: then "wiU 
the arcs AMD and ENG be equal. 

Draw the radii CD and OG. The triangles A CD 
and EOG haye all the sides of the one equal to the cor- 
responding sides of the other ; they are, therefore, equal in 
all their parts: hence, the angle A CD is equal to EOG. 
If, now, the sector A CD be placed upon the sector EOG, 
so that the an^le A CD shall coincide with the angle EOG, 
the sectors will coincide throughout ; and, consequently, the 
arcs AMD and ENG will coincide : hence, they will be 
equal ; which icas to be proved. 



PEOPOSITIOIT V. THEOREM. 

In equal circles, a greater arc is subtended by a greater 

chord ; and conversely, a greater chord subtends a greater 

arc. 

1°. In the equal circles J^— 

M//\ 
ADD and EGE, let the 

arc EGP be greater than 

the arc AMD : then will 

the chord EP be greater 

than the chord AD. 

For, place the circle EGK upon AHL, 60 that the cen- 
tre shall fall upon the centre C, and the point E upon 
A ; then, because the arc EGP is greater than AMD, the 
point P will fall at some point H, beyond D, and the 
chord EP will take the position AIL 

Draw the radii CA, CD, and CJI. Now, the sides 
AC, CJI, of the triangle ACII, are equal to the sides 
AC, CD, of the triangle A CD, and the angle ACJS is 




64: 



GEOMETRY. 



greater than A CD : hence, the side AH, or its equal EP y 
is greater than the side AD (B. I., P. IX.) ; which was to 
be proved. 

2°. Let the chord EP, 
W its equal AH, be great- 
er than AD : then will the 
arb EGP, or its equal 
ADH, be greater than 
AMD. 

For, if ADH were equal to AMD, the chord AH 
would be equal to the chord AD (P. IY.) ; which is con- 
trary to the hypothesis. And, if the arc ADH were less 
than AMD, the chord AH would be less than AD ; 
which is also contrary to the hypothesis. Then, since the 
arc ADH, subtended by the greater chord, can neither be 
equal to, nor less than AMD, it must be greater than 
AMD ; which was to be proved. 




PROPOSITION VI. THEOREM. 

The radius which is perpendicular to a chord, bisects that 
chord, and also the arc subtended by it. 

Let CG be the radius which is 
perpendicular to the chord AB : 
then will tin's radius bisect the chord 
AD, and also the arc AGP. 

For, draw the radii CA and CB. 
Then, the right-angled triangles CD A 
and CDB will have the hypothenuse 
CA equal to CB, and the side CD 
common ; the triangles are, therefore, equal in all their 
parts : hence, AD is equal to DB. Again, because CG 




BOOK III. 65 

is perpendicular to AB, at its middle point, the chord? 
GA and GB are equal (B. I., P. XVI.) ; and consequently, 
the arcs GA and GB are also equal (P. IV.) ■ hence, CG 
bisects the chord AB, and also the arc AGB ; which was 
to be proved. 

Cor. A straight line, perpendicular to a chord, at its mid 
die point, passes through the centre of the circle. 

Scholium. The centre C, the middle point D of the 
chord AB, and the middle point G of the subtended arc, 
are points of the radius perpendicular to the chord. But 
two points determine the position of a straight line (A. 11): 
hence, any straight line which passes through two of these 
points, will pass through the third, and be perpendicular to 
the chord. 



PROPOSITION VII. THEOREM. 

Tlirough any three points, not in the same straight line, one 
circumference may be made to pass, and but one. 

Let A, B, and C, be any three points, not in a 
straight line: then may one circumference be made to pass 
through them, and but one. 

Join the points by the lines 
AB, BC, and bisect these lines 
by perpendiculars DE and FG : 
then will these perpendiculars 
meet in some point 0. For, 
if they do not meet, they are 
parallel ; and if they are parallel, 

the line ABK, which is perpendicular to DE, is also per- 
pendicular to KG (B. I., P. XX., C. 1) ; consequently, there 
are two lines BE and BE, drawn through the same 

5 




66 



GEOMETRY. 




point B, and perpendicular to the same line KG ; which 
is impossible : hence, BE and EG meet in some point 0. 

Now, O is on a perpendicu- 
lar to AB at its middle point, 
it is, therefore, equally distant 
from A and B (B. L, P. XVL). 
For a like reason, O is eaually 
distant from B and C. If, 
therefore, a circumference be de- 
scribed from O as a centre, -with a radius equal to OA, 
it will pass through A, B, and G. 

Again, is the only point which is equally distant from 
A, B, and G : for, BE contains all of the points which 
are equally distant from A and B\ and EG all of the 
points which are equally distant from B and G ; and con- 
sequently, their point of intersection O, is the only point 
that is equally distant from A, B, and G : hence, one 
circumference may be made to pass through these points, and 
but one ; which was to be proved. 

Gor. Two circumferences cannot intersect in more than 
two points ; for, if they could intersect in three points, there 
would be two circumferences passing through the same three 
points ; which is impossible. 



PROPOSITION VIII. THEOEEM. 

In equal circles, equal chords are equally distant from the 
centres ; and of two unequal chords, the less is at the 
greater distance from the centre. 

1°. In the equal circles AG H and KLG, let th 
chords A G and KL be equal : then will they be equally 
distant from the centres. 



BOOK III. 



67 




For, let the circle KLG be placed upon A CRT, so that 
the centre R shall fall upon the centre 6>, and the point 
K upon the point A : 
then will the chord KL 
coincide with A C (P. 
[V.) ; and consequently, 
they will be equally dis- 
tant from the centre ; 
which icas to be proved. 

2°. Let AB be less than KL : then will it be at a 
greater distance from the centre. 

For, place the circle KLG upon ACH, so that R ; 
shall fall upon 0, and K upon A. Then, because the 
chord KL is greater than AB, the arc KSL is greater 
than AMB ; and consequently, the point L will fall at a 
point (7, beyond B, and the chord KL will take the 
direction A C. 

Draw OL> and OE, respectively perpendicular to AC 
and AB ; then will OE be greater than OF (A. 8), and 
OF than <9Z> (B. I., P. XV.) : hence, OE is greater than 
OB. But, OE and 0Z> are the distances of the two 
chords from the centre (B. I., P. XV., C. 1) : hence, the less 
chord is at the greater distance from the centre ; whieh was 
to be proved. 

Scholium. All the propositions relating to chords and arcs 
of equal circles, are also true for chords and arcs of one and 
the same circle. For, any circle may be regarded as made 
up of two equal circles, so placed, that they coincide in all 
their parts. 



68 



GEOMETRY. 




PKOPOSITION IX. THEOEEM. 

If a straight line is perpendicular to a radius at its essr 
tremity, it will be tangent to the circle at that point; 
conversely, if a straight line is tangent to a circle at 
any point, it will be perpendicular to the radius drawn 
to that point. 

1°. Let BD be perpendicular to the radius CA y at 
A : then will it be tangent to the circle at A. 

For, take any other point of 
JBD, as E, and draw GE : 
then will GE be greater than 
GA (B. L, P. XV.) ; and con- 
sequently, the point E will lie 
without the circle : hence, BD 
touches the circumference at the 

point A ; it is, therefore, tangent to it at that point (D. 11); 
which was to be proved. 

2°. Let BD be tangent to the circle at A : then will 
it be perpendicular to GA. 

For, let E be any point of the tangent, except the 
point gf contact, and draw GE. Then, because BD is a 
tano-ent, E lies without the circle ; and consequently, GE 
is greater than GA : hence, GA is shorter than any other 
line that can be drawn from G to BD ; it is, therefore, 
perpendicular to BD (B. L, P. XV., C. 1) ; which was to 
be proved. 

Gor. At a given point of a circumference, only one tan- 
gent can be drawn. For, if two tangents could be drawn, 
they would both be perpendicular to the same radius at the 
same point ; which is impossible (B. I., P. XIV.). 



BOOK III. 



69 



PEOPOSITIO^ X. THEOKEM. 



Two parallels 'biter cept equal arcs of a circumference. 



There may be three cases : both parallels may be secants ; 
one may be a secant and the other a tangent ; or, both 
may be tangents. 

1°. Let the secants AB and DJE be parallel : then 
will the intercepted arcs 3IJST and PQ be equal. 

For, draw the radius CM 
perpendicular to the chord 
MP ; it will also be per- 
pendicular to NQ (B. I., P. 
XX., C. 1), and M will be at 
the middle point of the arc 
J/ZTP, and also of the arc 
NIIQ : hence, MJ5F, which is 
the difference of MN and II3I, 

is equal to PQ, which is the difference of MQ and HP 
(A. 3) ; which was to be proved. 




2°. Let the secant AB and tangent DM, be parallel 
then will the intercepted arcs 3III and PH be equal. 

For, draw the radius CII 
to the point of contact II ; 
*it will be perpendicular to DJE 
(P. IX.), and also to its par- 
allel MP. But, because CII 
is perpendicular to J/1P, H 
is the middle point of the arc 
MIIP ( P. VI.) : hence, MH 
and PII are equal ; which 
was to be proved. 




70 



GEOMETRY. 



3°. Let the tangents BE and IB be parallel, and let 
H and E be their points of contact : then will the in- 
tercepted arcs EME and EPE be equal. 

For, draw the secant AB 
parallel to BE ; then, from 
what has just been shown, we 
shall have EM equal to EP, 
and ME equal to PE ": hence, 
EME, which is the sum of 
EM and ME, is equal to 
EPE, which is the sum of 
HP and PE; which was to 
be proved. 




PEOPOSITION XI. THEOKEM. 

If two circumferences intersect each other, the points of in- 
tersection loill be in a perpendicular to the line joining 
their centres, and at equal distances from it. 



Let the circumferences, whose centres are C and D, 
intersect at the points A and 
B : then will CD be perpen- 
dicular to AB, and AF will 
be equal to BE. 

For, the points A and B, 
being on the circumference 
whose centre is C, are equally- 
distant from C ; and being on 

the circumference whose centre is P, they arc equally dis- 
tant from B : hence, CB is perpendicular to A B at its 
middle point (B. I., P. XVI., C.) ; which teas to be proved. 




BOOK III. 



71 



PROPOSITION Xn. THEOREM. 

» 

If two circumferences intersect each other, the distance be- 
tween their centres will be less than the sum, and greater' 
than the difference, of their radii. 

Let the circumferences, whose centres are C and D, 
intersect at A : then will CD 
be less than the sum, and 
greater than the difference of 
the radii of the two circles. 

For, draw AC and AD, 
forming the triangle ACD. 
Then will CD be less than 
the sum of AC and AD, 

and greater than their difference (B. L, P. YIT.) ; which was 
to be proved. 




PROPOSITION XIII. THEOREM. 

If the distance between the centres of tioo circles is equal 
to the sum of their radii, they will be tangent externally. 

Let C and D be the centres of two circles, and let 
the distance between the centres be equal to the sum of the 
radii : then will the circles be tangent externally. 

For, they will have a point 
A, on the line CD, common, 
and they will have no other 
point in common ; for, if they 
had two points in common, the 
distance between their centres 
would be less than the sum of 
their radii ; which is contrary to the hypothesis 
are tangent externally ; which was to be proved. 




hence, they 



72 



GEOMETRY. 




PEOPOSITION XIV. THEOEEM. 

If the distance between the centres of two circles is equal to 
the difference of their radii, one will be tangent to the 
other internally. 

Let C and D be the centres of two circles, and let 
the distance between these centres be equal to the difference 
of the radii : then will the one be tangent to the other in- 
ternally. 

For, they will have a point A, on 
DC, common, and they will have no 
other point in common. For, if they 
had two points in common, the distance 
between their centres would be greater 
than the difference of their radii ; 
which is contrary to the hypothesis : 

hence, one touches the other internally ; which was to be 
proved. 

Cor. 1. If two circles are tangent, either externally or 
internally, the point of contact will be on the straight line 
drawn through their centres. 

Cor. 2. All circles whose centres are on the same straight 
line, and which pass through a common point of that line, 
are tangent to each other at that point. And if a straight 
line be drawn tangent to one of the circles at their common 
point, it will be tangent to them all at that point. 

Scholium. From the preceding propositions, we infer that 
two circles may have any one of six positions with respect 
to each other, depending upon the distance between their 

centres : 

1°. When the distance between their centres is greater 



BOOK J II. 



73 



than the sum of their radii, they are external, one to the 
other : 

2°. When this distance is equal to the sum of the radii, 
they are tangent, externally: 

3°. When this distance is less than the sum, and greater 
thap the difference of the radii, they intersect each other : 

4°. When this distance is equal to the difference of then 
radii, one is tangent to the other, internally: 

5°. When this distance is less than the difference of the 
radii, one is wholly within the other : 

6°. When this distance is equal to zero, they have a 
common centre; or, they are concentric. 



PROPOSITION XV. THEOREM. 

In equal circles, radii making equal angles at the centre, 
intercept equal arcs of the circumference ; conversely, 
radii which intercept equal arcs, make equal angles at the 
centre. 

1°. In the equal circles ADD and EGF, let the an- 
gles A CD and EOG be equal: then will the arcs AMD 
and EJSFG be equal. 

For, draw the chords AD 
and EG ; then will the tri- 
angles A CD and EOG have 
two sides and their ^included 
angle, in the one, equal to 
two sides and their included 

angle, in the other, each to each. They are, therefore, equal 
in all their parts ; consequently, AD is equal to EG. 
But, if the chords AD and EG are equal, the arcs AMD 
and ENG are also equal (P. IV.) ; which was to be proved. 




74: 



GEOMETRY. 



2°. Let the arcs AMD and EJSFG be equal : then will 
the angles A CD and EOG be equal. 

For, if the arcs AMD 
and ENG are equal, the 
chords AD and EG are 
equal (P. IV.) ; consequently, 
the triangles A CD and EOG 
have their sides equal, each 
to each ; they are, therefore, 

equal in all their parts : hence, the angle A CD is equal 
to the angle EOG ; which was to he proved. 




PROPOSITION XVI. THEOREM. 

In equal circles, commensurable angles at the centre are pro- 
portional to their intercepted arcs. 

In the equal circles, whose centres are C and 0, let > 

the angles A CD and DOE be commensurable; that is, 

let them have a common unit : then will they be propor- 
tional to the intercepted arcs AD and DE. 




P 7 



Let the angle if be a common unit ; and suppose, for 
example, that this unit is contained 7 times in the angle 
A CD, and 4 times in the angle DOE. Then, suppose 
A CD be divided into 7 angles, by the radii Cm, Cn, Cp, 
&c. ; and D OE into 4 angles, by the radii Ox, Oy, and 
Os, each equal to the unit M. 



BOOK ill. 75 

From the last proposition, the arcs Am, mn, &c, Bx, 
xy, &c, are equ# to each other ; and because there are 7 
of these arcs in AB, and 4 in BE, we shall have, 

arc AB : arc BE : : 7:4. 

Bat, by hypothesis, we have, 

angle A CB : angle B OE : : 7:4; 

hence, from (B. II., P. IV.), we have, 

anHe ACB : anG:le BOE : : arc AB : arc BE. 

If any other numbers than 7 and 4 had been used, the 
same proportion would have been found ; which was to be 
proved. 

Cor. If the intercepted arcs are commensurable, they will 
be proportional to the corresponding angles at the centre, 
as may be shown by changing the order of the couplets in 
the above proportion. 



PROPOSITION XVII. THEOREM. 

In equal circles, incommensurable angles are proportional to 

their intercepted arcs. 

In the equal circles, whoso 
centres are C and 0, let 
A CB and EOII be incom- 
mensurable : then will they 
be proportional to the arc3 
AB and FIL 

For, let the less angle FOIL, be placed upon the greater 
angle ACB, so that it shall take the position ACB. 




76 



GEOMETRY. 



Then, if. the proposition is not 
true, let us suppose that the 
angle A CB is to the angle 
FOII, or its equal A CD, 
as the arc AB is to an arc 
AO, greater than FH, or 
its equal AB ; whence, 




dT° 



angle A CB 



angle A CB 



arc AB : arc AO. 



Conceive the arc AB to be divided into equal parts, 
each less than DO : there will be at least one point of 
division between D and ; let I be that point ; and 
draw CI. Then the arcs AB, AT, will be commensura- 
ble, and we shall have (P. XVI.), 



angle A CB 



angle A CI 



arc AB 



arc AT, 



Comparing the two proportions, we see that the antecedents 
are the same in both : hence, the consequents are propor- 
tional (B. II., P. IV., C.) ; hence, 



angle A CB 



angle A CI 



arc A 



arc AI. 



But, AO is greater than AI : hence, if this proportion is 
true, the angle A CB must be greater than the angle A CI 
On the contrary, it is less : hence, the fourth term of the 
proportion cannot be greater than AB. 

In a similar manner, it may be shown that the fourth 
term cannot be less than AB : hence, it must be equal to 
AD ; therefore, we have, 



angle A CB 



angle A CB 



arc AB 



arc AB 



which was to be proved. 



Cor. 1. The intercepted arcs are proportional to the cor- 



BOOK III. 77 

responding angles at the centre, as may be shown by chang- 
ing the order of the couplets in the preceding proportion. 

Cor, 2. In equal circles, angles at the centre are pro- 
portional to their intercepted arcs ; and the reverse, whether 
they are commensurable or incommensurable. 

Cor 3. In equal circles, sectors are proportional to their 
angles, and also to their arcs. 

Scholium. Since the intercepted arcs are proportional to 
the corresponding angles at the centre, the arcs may be 
taken as the measures of the angles. That is, if a circum- 
ference be described from the vertex of any angle, as a cen- 
tre, and with a fixed radius, the arc intercepted between the 
sides of the angle may be taken as the measure of the 
angle. In Geometry, the right angle which is measured by 
a quarter of a circumference, or a quadrant, is taken as a 
unit. If, therefore, any angle be measured by one-half or 
two-thirds of a quadrant, it will be equal to one-half or 
two-thirds of a rio-ht an^le. 



PROPOSITION XVUI. THEOREM. 

An inscribed angle, is measured by half of the arc included 

between its sides. 

There may be three cases : the centre of the circle may 
lie on one of the sides of the angle ; it 
may lie within the angle ; or", it may ^^A. 

lie without the anirle. 

1°. Let EAD be an inscribed an- 
gle, one of whose sides AE passes 
through the centre : then will it be 
measured by half of the arc DE. 




78 



GEOMETRY. 



For, draw the radius CD. The external angle DCE, 
of the triangle DC A, is equal to the sum of the opposite 
interior angles CAD and CDA (B. I., P. XXV., C. 6). 
But, the triangle DC A being isosceles, 
the angles D and A are equal ; 
therefore, the angle DCE is double 
the angle DAE. Because DCE is 
at the centre, it is measured by the 
arc DE (P. XVII., S.) : hence, the, 
angle DAE ■ is measured by half of 
the are DE ; which was to be proved. 




2°. Let DAB be an inscribed angle, and let the centre 
lie within it : then will the angle be measured by half of 
the arc BED. 

For, draw the diameter AE. Then, from what has just 
been proved, the angle DAE is measured by half of DE, 
and the angle EAB by half of EB : hence, BAD, which 
is the sum of EAB and DAE, is measured by half of 
the sum of DE and EB, or by half of BED ; which 
was to be proved. 

3°. Let BAD be an inscribed angle, and let the centre 
lie without it : then will it be measured by half of the arc 
arc BD. 

For, draw the diameter AE. Then, 
from what precedes, the angle DAE 
is measured by half of DE, and the 
angle BAE by half of BE : hence, 
BAD, which is the difference of BAE 
and DAE, is measured by half of the 
difference of BE and DE, or by 
half of the arc BD ; which zoas to be proved. 




L> E 



BOOK III. 



79 



Cor. 1. All the angles BAG, 
BDC, BEC, inscribed in the same 
segment, are equal ; because they are 
each measured by half of the same 
arc BOC. 



Cor. 2. Any angle BAD, in- 
scribed in a semi-circle, is a, right an- 
gle ; because it is measured by half 
the semi-circumference BOB, or by 
a quadrant (P. XVII., S.). 




Cor. 3. Any angle BAG, in- 
scribed in a segment greater than a 
semi-circle, is acute ; for it is mea- 
sured by half the arc BOG, less 
than a semi-circumference. 

Any angle BOC, inscribed in a 
segment less than a semi-circle, is 
obtuse ; for it is measured by half the arc BA C, greater 
than a semi-circumference. 




Cor. 4. The opposite angles A 
and C, of an inscribed quadrilateral 
ABGB, are together equal to two 
right angles ; for the angle DAB 
is measured by half the arc DCB, 
the angle DCB by half the arc 

DAB : hence, the two angles, taken together, are mea- 
sured by half the circumference : hence, their sum is equal 
to two right angles. 




80 



GEOMETRY. 



PROPOSITION XIX. THEOREM. 

Any angle formed by two chords, which intersect, is mea- 
sured by half the sum of the included arcs. 

\ Let DEB be an angle formed by the intersection of 
the chords AB and CD : then will it be measured by 
half the sum of the arcs A C and DB. 

For, draw AF parallel to DC : 
then, the arc DF will be equal to 
AC (P. X.), and the angle FAB f 
equal to the angle DEB (B. I., P. 
XX., C. 3). But the angle FAB is 
measured by half the arc FDB (P. 
XVIIL) ; therefore, DEB is measured 

by half of FDB; that is, by half the sum of FD and 
DB, or by half the sum of AC and DB ; which was to 
be pi % oved. 




PROPOSITION XX. 



THEOREM. 



The angle formed by two secants^ is measured by half the 

difference of the included arcs. 

Let AB, AC, be two secants : then will the angle 
BAG be measured by half the differ- 
ence of the arcs BG and DF. 

Draw DE parallel to AC : the 
arc EC will be equal to DF (P. X.), 
and the angle BDE equal to the an- 
gle BAG (B. I., P. XX., C. 3.). But 
BDE is measured by half the arc 
BE (P. XVIIL) : hence, BA C is 
also measured by half the arc BE ; 
that is, by half the difference of BG 
and EC, or by half the difference of BG and DF\ which 
was to be proved. 




BOOK III. 



81 



PROPOSITION XXI. THEOEEM. 

An angle formed by a tangent and a chord meeting it at 
the point of contact, is measured by half the included. 
arc. 



Let BE be tangent to the circle AMC, and let AG 
be a chord drawn from the point of contact A : then 
will the angle BAG be measured 
by half of the arc AMC. 

For, draw the diameter AD. 
The angle BAD is a right an^le 
(P. IX.), and is measured by half 
the semi-circumference AMD (P. 
XVn., S.) ; the angle DAG is 
measured by half of the arc DG 
(P. XVIII.) : hence, the angle BAG, 

which is equal to the sum of the angles BAD and DAG, 
is measured by half the sum of the arcs AMD and DG 
or by half of the arc AMC ; which was to be proved. 

The angle CAE, which is the difference of DAE and DAG, 
is measured by half the difference of the arcs DC A and DC, 
or by half the arc CA. 




PEACTICAL APPLICATIONS. 



PROBLEM I. 



To bisect a given straight line. 

Let AB be a given straight line. 

From A and B, as centres, with 
a radius greater than one half of AB, 
describe arcs intersecting at E and 
F: join E and F, by the straight 
line EF. Then will EF bisect the 
given line AB. For, E and F 
are each equally distant from A and 
B ; and consequently, the line EF 
bisects AB (B. I., P. XYI., C.). 



A 



X E 



C 



B 



PROBLEM H. 



To erect a perpendicular to a given straight line, at a given 

point of that line. 



y® 



Let EF be a given line, and let A be a given point on 

that line. 

From A, lay off the equal 
distances AB and'^LC; from 

B and (7, as centres, with a 

radius greater than one half E B A C 



BOOK III. 



83 



of DC, describe aics intersecting at D ; draw tie line AD: 
then will AD be the perpendicular required. For, D and A 
are each equally distant from D and C ; consequently, DA is 
perpendicular to DO (B. L, P. XVI, C). 



PEOBLEM III. 

« 
To draw a perpendicular to a given straight line, from a 

given point without that line. 

Let DD be the given line, and A the given point. 

From A, as a centre, with a ra- 
dius sufficiently great, describe an arc 
cutting DD in two points, D and 
D ; with D and D as centres, and 
a radius greater than one-half of DD, 
describe arcs intersecting at E; draw 
AE : then will AE be the perpendi- 
cular required. For, A and E are each equally distant 
from D and D : hence, AE is perpendicular to DD 
(B. I., F. XVI, C). 



F 



C 



E 



^'D 



PEOBLEM rV. 

At a point on a given line, to construct an angle equal to 

a given angle. 

Let A be the given point, AD the given line, and 
IKD the given angle. 

From the vertex K as a 
centre, with any radius KI, 

describe the arc ID. tcrminat- 

JV " i — A B 

mg in the sides of the angle. 

From A as a centre, with a radius AD, equal to KI y 




84 



GEOMETRY. 



describe the indefinite arc BO ; then, with a radius equal 
to the chord BI, from 5 as a centre, describe an arc 
cutting the arc BO in D ; 
draw AB : then will BAB 
be equal to the angle K. 

For, the arcs BB, IB, 
have equal radii and equal 

chords : hence, they are equal (P. IV.) ; therefore, the angles 
BAB, 1KB, measured by them, are also equal (P. XV.). 




I A 



PEOBLEM Y. 
To bisect a given arc, or a given angle. 

1°. Let ABB be a given arc, and C its centre. 

Draw the chord AB ; through C, 
draw CB perpendicular to AB (Prob. 
HI.) : then will CB bisect the arc 
ABB (P. VI.) . 




2°. Let A CB be a given angle. A /^ 

Vv r ith C as a centre, and any 
radius CB, describe the arc BA ; 
1 ' cct it by the line CB, as« just 
explained : then will CB bisect the angle ACB. 

For, the arcs AE and EB are equal, from what was 
just shown; consequently, the angles ACE and ECB are 
also equal (P. XV.). 

Scholium. If each half of an arc or angle be bisected, 
the original arc or angle will be divided into four equal 
parte; and if each of these be bisected, the original arc or 
angle will be divided into eight equal parts ; and so on. 



BOOK III. 



£5 



PEOBLEM VI. 

Through a given point, to draw a line parallel to a given 

line. 

Let A be a given point, and BC a given line. 

From the point A as a centre, 
with a radius AE, greater than the 
shortest distance from A to BC, 
describe an indefinite arc EG ; from 
^ as a centre, with the same ra- 
dius, describe the arc AF ; lay off 
ED equal to AF, and draw AD : then will AD be the 
parallel required. 

For, drawing AE, the angles AEF, FAD, are equal 
(P. XV.) ; therefore, the lines AD, EF are parallel (B. L, 
P. XIX., C. 1.). 




L 




PEOBLE}! TIL 

Given, two angles of a triangle, to construct the third 

angle. 

Let A and B be given angles of a triangle. 

Draw a line DF, and at some 

point of it, as E, construct the an- 

• gle FEII equal to A, and IIEC 

equal to B. Then, will CED be 

equal to the required angle. 

For, the sum of the three angles at E is equal to two 
light angles (B. L, P. L, C. 3), as is also the sum of the 
three angles of a triangle (B. L, P. XXV.). Consequently, 
the third angle CED must be equal to the third angle of 
the triangle. 




86 



GEOMETRY. 



PROBLEM VIII. 

Give?i, two sides and the included angle of a triangle, to 

construct the triangle. 

Let B and C denote the given sides, and A the given 
angle. 

Draw the indefinite line DE, 
and at D construct an angle 
EDE, equal to the angle A ; on 
DE, lay off DM equal to the 
side (7, and on DE, lay off 
DG equal to the side D ; draw 
GM: then will DGM be the required triangle (B. I, P. V.), 




L — F 



PROBLEM IX. 

Given, one side and two angles of a triangle, to construct 

the triangle. 

The two angles may be either both adjacent to the given 
side, or one may be adjacent and the other opposite to it. 
In the latter case, construct the third angle by Problem TIL 
We shall then have two angles and their included side. 

Draw a straight line, and on it 
lay off DE equal to the given 
side ; at D construct an angle 
equal to one of the adjacent an- 
gles, and at E construct an angle 
equal to the other adjacent angle ; 

produce the sides DE and EG till they intersect at M : 
then will DEM be the triangle required (B. L, P. VI.). 




BOOK III. 



PEOBLEM X. 



87 




Given, the three sides of a triangle, to construct the tri- 
angle. 

Let A, B, and C, be the given sides. 

Draw DE, and make it equal 
to the side A ; from D as a 
centre, with a radius equal to the 
side -Z?, describe an arc ; from E 
as a centre, with a radius equal 
to the side C, describe an arc 

intersecting the former at F ; draw DF and EF : then 
will DEF be the triangle required (B. I, P. X.). 

Scliplhim. In order that the construction may be possible, 
any one of the given sides must be less than the sum of the 
other two, and greater than their difference (B. I, P. VII., S.). 

PEOBLEM XI. 

Given, two sides of a triangle, and the angle opposite one 
of them, to construct the triangle. 

Let A and B be the given sides, and C the given 
angle. 

Draw an indefinite line DG, A; 

and at some point of it, as Z>, r> 

construct an angle GDE equal 

to the given angle ; on one side 

of this angle lay off the distance 

BE equal to the side B adjacent / 

to the given angle ; from E as 

a centre, with a radius equal to the side opposite the given 

angle, describe an arc cutting the side I)G at G ; draw 

EG. Then will BEG be the req ' triangle. 




83 GEOMETRY. 

For, the sides BE and EG are e|ual to the given 
sides, and the angle B, opposite one of them, is equal to 
the given angle. 

Scholium, When the side opposite the given angle is 
greater than the other given side, there will he hut one 
solution. When the given angle is acute, and the side 
opposite the given angle is less 
than the other given side, and 
greater than the shortest dis- -^ 

tance from E to BG, there ^/^^\ 

will be two solutions, BEG j^ ^^\S _. >^- 

and BEE When the side £ X., ..-*' 

opposite the given angle is 

equal to the shortest distance from E to BG, the arc 
will he tangent to BG, the angle opposite BE will be 
a right angle, and there will be but one solution. When 
the side opposite the given angle is shorter than the distance 
from E to BG, there will be no solution. 

PROBLEM XII. 

Given, two adjacent sides of a parallelogram and their 
included angle, to construct the parallelogram. 

Let A and B be the given sides, and G the given 

angle. 

Draw the line BIT, and 
at some point as B, construct 
the angle EBF equal to the 

angle C. Lay off BE equal D^ ^ ~H 

to the side A, and BE equal ^ . , / 

to the side B ; draw EG B , , & . 

parallel to BE, and EG par- 
allel to BE- then will BEGE be the parallelogram re. 

quired. 





BOOK III. 



89 



For, the opposite sides are parallel by construction ; and 
consequently, the figure is a parallelogram (D. 28) ; it is 
also formed with the given sides and given angle. 



PROBLEM xirr. 
To find the centre of a given circumference. 

Take any three points A, 
J5 y and C, on the circumference , 
or arc, and join thern by the 
chords AB, JB C ; bisect these 
chords by the perpendiculars DJE 
and FG : then will their point, 
of intersection 0, be the centre 
required (P. VII.). 

Scholium. The same construc- 
tion enables us to pass a circumference through any three 
points 'not in a straight line. If the points are vertices of 
a triangle, the circle will be circumscribed about it. 




PROBLEM XIV. 
Through a given point, to draw a tangent to a given circle. 

There may be two cases : the given point may lie on 
the circumference of the given circle, or it may lie without 
the given circle. 

1°. Let C be the centre of the 
given circle, and A a point on the 
circumference, through which the tan- 
gent is to be drawn. 

Draw the radius CA, and at A 
draw AD perpendicular to AG: then 
will AD be the tangent required (P. IX.). 




90 



GEOMETRY. 




2°. Let C be tlie centre of the given circle, and A a 
point without the circle, through which the tangent is to be 
drawn. 

Draw the line A G ; bisect it at 
0, and from as a centre, with a 
radius 00, describe the circumference 
ABCD; join the point A with the 
points of intersection D and B : 
then will both AD and AB be 
tangent to the given circle, and there 
will be two solutions. 

For, the angles ABO and ABO 
are right angles (P. XVIII., C. 2) : 

hence, each of the lines AB and AD is perpendicular to 
a radius at its extremity ; and consequently, they are tangent 
to the given circle (P. IX.). 

Corollary. The right-angled triangles ABO and ADC, 
have a common hypothenuse A C, and the side B O equal 
to DO; and consequently, they are equal in all their parts 
(B. I., P. XVII.) : hence, AB is equal to AD, and 
the angle GAB is equal to the angle CAD, The tan- 
gents are therefore equal, and the line A C bisects the 
angle between them. 



PEOBLEM XV. 



To inscribe a circle in a given triangle. 

Let ABO be the given 
triangle. 

Bisect the angles A and 
B, by the lines AG and 
BO, meeting in the point O 
(Prob. V.) ; from the point 




BOOK III. 



91 



let fall the perpendiculars OD, OE, OF, on the sides of 
the triangle : these perpendiculars will all be equal. 

For, in the triangles BOD and BOE, the angles OBE 
and OBB are equal, by construction ; the angles ODB 
and OJEB are equal, because both are right angles ; and 
consequently, the angles BOB and BOE are also equal 
(B. I., P. XXV., C. 2), and the side OB is common ; and 
therefore, the triangles are equal in all their parts (B. I., 
P. VI.) : hence, OD is equal to OE. In like manner, it 
may be shown that OD is equal to OF. 

From as a centre, with a radius OD, describe a 
circle, and it will be the circle required. For, each side is 
perpendicular to a radius at its extremity, and is therefore 
tangent to the circle. 

Corollary. The lines that bisect the three angles of a 
triangle all meet in one point. 



PEOBLEM XVI. 

On a given line, to construct a segment that shall contain 

a given angle. 

Let AB be the given line. 





Produce AB towards D ; at B construct the angle 
DDE equal to the given angle draw BO perpendicular 



92 



GEOMETRY. 



to BE, and at the middle point G, of AB, draw GO 
perpendicular to AB ; from their point of intersection 0, 
as a centre, with a radius OB, describe the arc AMB : 
then will the segment AMB be the segment required. 





For, the angle ABE, equal to EBB, is measured by- 
half of the arc AKB (P. XXI.) ; and the inscribed angle 
AMB is measured by half of the same arc : hence, the 
angle AMB is equal to the angle EBB, and conse- 
quently, to the given angle. 



t 



BOOK IV. 

MEASUREMENT AND RELATION OF POLYGONS. 

DEFINITIONS. 

1. Similar Polygons, are polygons which are mutually 
equiangular, and which have the sides about the equal angles, 
taken in the same order, proportional. 

2. In similar polygons, the parts which are similarly 
placed in each, are called homologous. 

The corresponding rngles are homologous angles, the 
corresponding sides are homologous sides, the corresponding 
diagonals are homologous diagonals, and so on. 

3. Similar Arcs, Sectors, or Segments are those which 
correspond to equal angles at the centre. 

Thus, if the angles A and are 
equal, the arcs BFO and DGJ3 are 
similar, the sectors JBAC and DOJE 
are similar, and the segments BFO 
and DGE are similar. 

4. The Altitude of a Triangle, is the perpendicular 
distance from the vertex of either an- 
gle to tjie opposite side, or the opposite 
side produced. 

The vertex of the angle from which 
the distance is measured, is called the 
vertex of the triangle, and the opposite 
side, is called the base of the triangle. 






94 GEOMETRY. 

5. The Altitude of a Parallelogram, is the perpen- 
dicular distance between two opposite 

sides. 

These sides are called bases y one the 
upper, and the other, the lower base. 

6. The Altitude of a Trapezoid, is the perpendicular 
distance "between its parallel sides. 

These sides are called bases / one the 
upper, and the other, the lower base. 

7. The Ajrea of a Surface, is its numerical value 
expressed in terms of some other surface taken as a unit. 
The unit adopted is a square described on the linear unit, 
as a side. 

PROPOSITION I. THEOREM. 

Parallelograms which have equal bases and equal altitudes, 

are equal. 

Let the parallelograms AJBCD and EFGH have equal 
bases and equal altitudes : then will the parallelograms be 
equal. 

For, let them be so placed 
that their lower bases shall 
coincide ; then, because they 
have the same altitude, their 
upper bases will be in the 
same line BG, parallel to AB. 

The triangles BAIT and CBG, have, the sides AB and 
BO equal, because they are opposite sides of the parallel- 
ogram AG (B. I, P. XXVIII.) ; the sides AH and BG 
equal, because they are opposite sides of the parallelogram 
AG ; the angles BAIT and CBG equal, because their 




BOOK IV. 95 

sides are parallel and lie in the same direction (B. L, 
P. XXIV.) : hence, the triangles are equal (B. I., P. V.). 

If from the quadrilateral ABGJD, we take away the tri- 
angle DAE", there will remain the parallelogram AG- \ if 
from the same quadrilateral ABGD, we take away the tri- 
iriano-le CBG, there will remain the parallelogram AC: 
hence, the parallelogram AC is equal to the parallelogram 
EG (A. 3) ; which was to be proved. 



PKOPOSITIOISr II. THEOKEM. 

A triangle is equal to one-half of a parallelogram having 
an equal base and an equal altitude. 

Let the triangle ABC, and the parallelogram ABED, 
have equal bases and equal altitudes : then will the triangle 
be equal to one-half of the parallelogram. 

For, let them be so 
placed that the base of 
the triangle shall coin- 
cide with the lower base 
of the parallelogram ; 

then, because they have equal altitudes, the vertex of the 
triangle will lie in the upper base of the parallelogram, or 
in the prolongation of that base. 

From A, draw AE parallel to BC, forming the par- 
allelogram ABCE This parallelogram will be equal to 
the parallelogram ABED, from Proposition I. But the 
triangle ABC is equal to half of the parallelogram ABCE 
. (B. I, P. XXVIIL, C. 1) : hence, it is equal to .half of 
the parallelogram ABED (A. 1) ; wUch was to be proved 



D E F c c 




Cor. Triangles having equal bases and equal altitudes are 
equal, for they are halves of equal parallelograms. 



96 



GEOMETRY. 



PKOPOSITION III. THEOEEM. 

Rectangles having equal altitudes, are proportional to their 

bases. 

There may be two cases : the bases may be commensu- 
rable, or they may be incommensurable. 

1°. Let ABCB and HEFK, be two rectangles whose 
altitudes AD and HK are equal, and whose bases AB 
and HE are commensurable : then will the areas of the 
rectangles be proportional to their bases. 



D 



C 



A 



K 



B 



F 



H 



E 



Suppose that AB is to HE, as 7 is to 4. Conceive 
AB to be divided into 7 equal parts, and HE into 4 
equal parts, and at the points of division, let perpendiculars 
be drawn to AB and HE. Then will ABCB be divi- 
ded into 7, and HEFK into 4 rectangles, all of which will 
be equal, because they have equal bases and equal altitudes 
(P. I.) : hence, we have, 



ABCB 



HEFK 



4. 



But we have, by hypothesis, 

AB : HE 



4. 



From these proportions, we have (if. II., P. IV.), 
' ABCB : HEFK : : AB : HE. 

Had any other numbers than 7 and 4 been used, the same 
proportion would have been found ; which was to be proved, 



BOOK IV. 97 

2°, Let the bases of the rectangles be incommensurable : 

then will the rectangles be proportional to their bases. 

For, place the rectangle HEFK 

upon the rectangle ABCD, so that T> F K C 

it shall take the position AEFD. 
Then, if the rectangles are not pro- 
portional to their ? \es, let us sup- jf E~To B 

pose that 

ABCD : AEFD :: AB : AO; 

in which AO is greater than AE. Divide AB into 
equal parts, each less than OF ; at least one point of 
division, as 7J will fall between F and ; at this point, 
draw IK perpendicular to AB. Then, because AB and 
AI are commensurable, we shall have, from what has just 
been shown, 

ABCD : AIKD : : AB : AI. 

The above proportions have their antecedents the same 
in each ; hence (B. II., P. IV., C), 

AEFD : AIKD : : AO : AI. 

The rectangle AEFD is Jess than AIKD; and if the 
above proportion Avere true, the line AO would be less 
than AI ; whereas, it is greater. The fourth term of the 
proportion, therefore, cannot be greater than AE. In like 
manner, it may be shown that it cannot be less than AE ; 
consequently, it must bo equal to AE : hence, 

ABCD : AEFD : : AB AE ; 

which was to be proved. 

Cor. If rectangles have equal bases, they are to each 
other as their altitudes. 



/ 



98 



GEOMETRY. 



C 



PEOPOSITION IV. THEOEEM. 

Any two rectangles are to each other as the products of 

their bases and altitudes. 

Let ABCB and AEGF be two rectangles: then will 
ABCB be to AEGF, as AB x AB is to AE x AF 

For, place the rectangles so 
that the angles BAB and EAF 
shall be opposite or vertical ; 
then, produce the sides CB 
and GE till they meet in H. 

The rectangles ABCB and 
ABHE have the same altitude 
AB : hence (P. HI.), 

ABCB : ABHE : : AB : AE. 



H IL. 




The rectangles ABHE and AEGF 
altitude J.JS' : hence, 



ABHE 



AEGF 



AB 



have the same 



AF. 



Multiplying these proportions, term by term (B. II., P. 
XII.), and omitting the common factor ABHE (B. II., 
P. VH), we have, 

ABCB : AEGF :: AB x AB : AE X AF ; 

which was to be proved. fc 

Scholium 1. If we suppose AE and ^Li^ each to b 
equal to the linear unit, the rectangle AEGF will be th 
superficial unit, and we shall have, 



ABCB 



AB x AB 



i; 



BOOK IV. 



99 



ABGD = AB x AZ> : 

hence, the area of a rectangle is equal to the product of 
its base and altitude ; that is, the number of superficial 
units in the rectangle, is equal to the product of the number 
of linear units in its base by the number of linear units in 
its altitude. 

Scholium 2. The product of two lines is sometimes called 
the rectangle of the Hues, because the product is equal to 
the area of a rectangle constructed with the lines as sides. 



PROPOSITION Y. THEOEEM. 



The area of a parallelogram is equal to the product of its 

base and altitude. 

Let ABGD be a parallelogram, AB its base, and BE 
its altitude: then will the area of ABCJD be equal to 
AB x BE. 

For, construct the rectangle 
ABEE, having the same base 
and altitude : then will the rec- 
tangle be equal to the parallelo- 
gram (P. I.) ; but the area of the 
rectangle is equal to AB x BE: 

hence, the area of tho parallelogram is also equal to 
AB x BE; which was to be proved. 

Cor. Parallelograms are to each other as tho products 
of their bases and altitudes. If their altitudes are equal, 
they are to each other as their bases. If their bases are 
equal, they are to each other as their altitudes. 




wo 



GEOMETRY. 




PROPOSITION VI. THEOREM. 

The area of a triangle is equal to half the 'product of iU 

base and altitude. 

Let ABO be a triangle, BO its "base, and AD. its 
altitude : then will the area of the triangle be equal to 

\BO x AD, 

For, from O, draw OE 
parallel to BA, and from A, 
draw AE parallel to OB.- The 
area of the parallelogram BCEA 
is B x AD (P. V.) ; but the 
triangle ABO is half of the par- 
allelogram BCEA: hence, its area is equal to IBCxAD; 
which was to be proved. 

Cor. 1, Triangles are to each other, as the products of 
their bases and altitudes (B. II., P. YJL). If their alti- 
tudes are equal, they are to each other as their bases. If 
their bases are equal, they are to each other as their alti- 
tudes. 

Cor. 2. The area of a triangle is equal to half the pro- 
duct of its perimeter and the radius of the inscribed circle. 

For, let DEE be a circle 
inscribed in the triruigle ABO. 
Draw OD, OE, and OF, to 
the points of contact, and OA, 
OB, and 00, to the verti- 
ces. 

The area of 0B0 will be 
equal to \0E X BG ; the 
area of OA will be equal to £ OF x A ; and the area 




BOOK IV. 



101 



of OAB will be equal to \OD x AB ; and since OD, 
OE, and OF, are equal, the area of the triangle ABC 
(A. 9), will he equal to \OB {AB + BC + 04). 



PROPOSITION Til. THEOREM. 

The area of a trapezoid is equal to the product of its alti- 
tude and half the sum of its parallel sides. 

Let ABCB he a trapezoid, BE its altitude, and AB 
and DC its parallel sides: then will its area he equal to 
DJExUAB+BC). 

For, draw the diagonal AC, form- 
in o- the triangles ABC and A CD. 
The altitude of each of these trian- 
gles is equal to DE. The area of 
ABC is equal to \AB x DE (P. 
VI.) ; the area of A CD is equal to 

\DC X DE : hence, the area of the trapezoid, which is the 
sum of the /triangles, is equal to the sum of \AB X DE 
and \DC X DE, or to DE x \{AB +DC) ; which was 
to be proved. 





A E 



B 



PROPOSITION Yin. THEOREM. 

TVie square described on the sum of two lines is equal to 

the sum of the squares described on the lines, increu- 
» by twice the rectangle of the lines. 



Let AB and BC he two lines, 

and AC their sum: then will 



AC 2 = AB 2 + BC 2 + 2yJLC xi?C, 

On A C, construct the square 
ACDE ; from J?, draw BIT par- 



E 



n 



1 



D 

G 



J 



102 



GEOMETRY. 



allel to AE ; lay off AF equal to AB, and from 
F, draw FG- parallel to AG : then will IG and IS be 
each equal to BG ; and ZS and ZF, to AB. 

The square AGBF is composed 
of four parts. The part ABIF is E HP 

a square described on ^4J? ; the part 

IGBH is equal to a square described jj -*-! |q 

on BG ; the part x BGGI is equal 

to the rectangle of AB and i?(7 ; 

and the part FIHE is also equal to B C 

the rectangle of AB and BG : and 

because the whole is equal to the sum of all its parts (A. 9), 

we have, 

AG 2 = AB 2 + BG 2 + 2AB x BG ; 

tchich teas to be proved. 

Gov. If the lines AB and BG are equal, the four 

parts of the square on A G will also be equal : hence, the 

square described on a line is equal to four times the square 
described on half the line. 



PROPOSITION IX. THEOREM. 

The square described on the difference of two lines is equal 
to the sum of the squares described on the lines, dimin- 
ished by twice the rectangle of the lines. 

Let AB and BG be two lines, and AG their differ- 
ence : then will 

AG 2 = AB 2 + BG 2 - 2AB x BG. 

On AB consti.net the square ABIF; from G draw 
GG parallel to B I ; lay off GD equal to AC, and 
from D draw DK parallel and equal to JjA ; complete 



BOOK IV. 



103 



L 
K 1 



E 



G 



D 



A 



C B 



the square EFLK : then will EK be equal to B (7, and 
EFLK will be equal to the square of BC. 

The whole figure ABILKE is 
equal to the sum of the squares 
described on AB and BC. The 
part CBIG is equal to the rect- 
angle of AB and BC ; the part 
BGLK is also equal to the rect- 
angle of AB and BC. If from 

the whole figure ABILKE, the two parts CBIG and 
BGLK be taken, there will remain the part ACDE, 
which is equal to the square of AC : hence, 

AC 2 = AB 2 + BC 2 - 2AB x BC ; 
which was to be proved. \jz. 









PROPOSITION X. 



THEOREM. 



F 



G I 



TVie rectangle contained by the sum and difference of two 
lines, is equal to the difference of their squares. 

Let AB and BC be two lines, of which AB is the 
greater : then will 

(AB + BC) {AB -BC) = AB 2 - BC 2 - 

On AB, construct the square 
ABIE ; prolong AB, and make 
BE equal to BC; then will AK E 

be equal to AB + B C ; from 
iT, draw KL parallel to BI, and 
make it equal to A C ; draw IE 
parallel to KA, and CG parallel 
to BI : then I)G is equal to 

BC, and the figure DIIIG is equal to the square on 
BC, and EBGF is equal to BKL1I. 



1) 



n 



iL 



B K 



104: 



GEOMETRY. 



F 



E 



G I 



II 



D 



If we add to the figure ABUE, the rectangle BKLH, 

we shall have the rectangle AJZLE, which is equal to the 

the rectangle of AB 4- BC and 

AB — BC. If to the same figure 

ABUE, we add the rectangle 

DGFE, equal to BKLH, we 

shall have the figure ABHBGF, 

which is equal to the difference of 

the squares of AB and BC. But 

the sums of equals are equal (A. 2), 

hence, 

{AB + BC) {AB -BC) = AB 2 - BC 2 1 

which was to be proved. 



C B K 



PEOPOSITION XI. THEOEEM. 



The square described on the hypothenuse of a right-angled 
triangle, is equal to the sum of the squares described on 
the other two sides. 

Let ABC be a triangle, right-angled at A : then will 
BC 2 = AB 2 + AC 2 . 

Construct the square BC on tie side BC, the square 
All on the side AB, and 
the* square AI on the side 
AC ; from A draw AD 
perpendicular to BC, and 
prolong it to E : then will 
BE be parallel to BE; 
draw AF and IIC. 

In the triangles HBC 
and ABE, we have JIB 
equal to AB, because they 
are sides of the same square j 




BOOK IV. 105 

BC equal to BJB\ for the same reason, and the included 
angles HBO and ABF equal, because each is equal to the 
angle ABO plus a right angle : hence, the triangles are 
equal in all their parts (B. I, P. V.). 

The triaifgle ABF, and the rectangle BE, hare the 
same base BE, and because DE is the prolongation of 
DA, their altitudes are equal : hence, the triangle ABF 
is equal to half the rectangle BE (P. IX). The triangle 
HBO, and the square BB, have the same base BIT, and 
because AC is the prolongation of AL (B. I., P. JV.V 
their altitudes are equal: hence, the triangle HBC is equal 
to half the square of AH. But, the triangles ABF and 
HBO are equal: hence, the rectangle BE is equal to the 
square AH. In the same manner, it may be shown that 
the rectangle BG is equal to the square AI : hence, the 
sum of the rectangles BE and BG, or the square BG, 
is equal to the sum of the squares AH and AI ; or, 
BC = 4-B + AC ; which was to be proved. 

Cor. 1. The square of either side about the rio-ht an^le 
is equal to the square of the hypothenuse diminished by the 
square of the other side : thus, 

AB 2 = BO 2 - AC 2 ; or, AC 2 = BC 2 - AB 2 . 

Cor. 2. If from the vertex of the right angle, a per- 
pendicular be drawn to the hypothenuse, dividing it into two 
segments, BB and BC, the square of the hypothenuse will 
be to the square of cither of the other sides, as the hypo- 
thenuse is to the segment adjacent to that side. 

For, the square BG, is to the rectangle BE, as BC 
to BB (P. III.) ; but the rectangle BE is equal to the 
square An : hence, 

BC 2 : AB 2 : : BC : BB. 



i06 



GEOMETRY. 



In like manner, we have, 



BC 2 : AC" : : BG : DC. 

Cor. 3. The squares of the sides about the right angle 
are to each other as the adjacent 
segments of the hypothenuse. A 



For, by combining the propor- 
tions of the preceding corollary 
(B. H., P. IV., C), we have, 




AJBT 



AC 



BD 



DC. 



AC 2 = 2AB 2 ; or, AC' = 2BC\ 



H D 




G 



Cor. 4. The square described on the diagonal of a 
square is double the given square. 

For, the square of the diagonal is 
equal to the sum of the squares of the 
two sides ; but the square of each side 
is equal to the given square : hence, 



C 



B F 



Cor. 5. From the last corollary, we have, 



AG" : AB 



-2 



2:1; 



hence, by extracting the square root of each term, we have, 

AC : AB : : i/2 : 1 ; 

that is, the diagonal of a square is to the side, as the 
square root of two to one / consequently, the diagonal and 
the side of a square are incommensurable. 



BOOK IV. 



107 



PEOEOSITIOX XII. THEOEEM. 

In any triangle, the square of a side opposite an acuta 
angle, is equal to the sum of the squares of the base and 
the other side, diminished by twice the rectangle of the 
base and the distance from the vertex of the acute angle 
to the foot of the perpendicular drawn from the vertex 
of the opposite angle to the base, or to the base produced. 



Let ABC be a triangle, C one 
of its acute angles, BC its base, and 
AD the perpendicular drawn from A 
to BC y or BO produced; then will 



■;- 




AB' = BC + AC 2 - 2BC x CD. 



For, whether the perpendicular meets the base, or tiie 
base produced, we have BD equal to the difference of 
BC and CD : hence (P. IX.), 

BD 2 = BC 2 + CD 2 -2BC x CD. A 

Adding AD' to both members, we 

have, 

D B 




BD 2 + AD 2 = BC 2 + CD 2 + AD 2 - 2BC x CD. 



But, BD 2 + AD 2 = AB\ and CD 2 + AD 2 = AC 2 : 

hence, 

AB 2 = BC 2 + AC 2 - 2BC x CD ; 

which was to be proved. 



108 GEOMETRY. 



PROPOSITION XIII. THEOREM. 

In any obtuse-angled triangle, the square of the side opposite 
the obtuse angle is equal to the sum of the squares of 
the base and the other side, increased by twice the rect- 
angle of the base and the distance from the vertex of the 
obtuse angle to the foot of the perpendicular drawn froin 
the vertex of the opposite angle to the base produced. 

Let AB be an obtuse-angled triangle, B its obtuse 
angle, BG its base, and AD the perpendicular drawn 
from A to BG produced ; then will 

AC 2 = BG 2 + AB 2 + 2JSC x BD. 

For, CD is the sum of BG A 
and BD : hence (P. VHL), 

GD 2 = BG 2 + BD 2 + 2BG x BD. 

Adding AD 2 to both members, E> & 

and reducing, we have, 

AG 2 - BG 2 + AB 2 -f 2BG x BD ; 
which was to be proved. N 

Scholium. The right-angled triangle is the only one in 
which the sum of the squares described on two sides is 
equal to the square described on the third side. 

PROPOSITION XIV. THEOREM. 

In any triangle, the sum of the squares described on two 
sides is equal to twice the square of half the third side 
increased by twice the square of the line drawn from 
the middle point of that side to the vertex of the opposite 
angle. 
Let ABG be any triangle, and JSA a line drawn from 



BOOK IV. 109 

the middle of the base BC to the vertex A : then will 



AB 2 + AC 2 = 2 BE 2 + 2JE4 1 . 

Draw ^1Z> perpendicular to J?(7; then, from Proposition 
XII., we have, 




4(? = #<? + JE4' - 2^(7 x ED. 

From Proposition XIII., we have, 

IB 2 = 2^ 2 + £l 2 + 2i?^ x ED. * ^ ^ 

Adding these equations, member to member (A. 2), recollect* 
ing that BE is equal to EC, we have, 

AD" + Z(7 2 = 2^ 2 + 2^[ 2 ; 
which was to he proved. 

Cor. Let ABCD be a parallelogram, and I?2>, ^4(7, 
its diagonals. Then, since the diagonals 
mutually bisect each other (B. I., P. 
XXXI. ), we shall have, 

AB 2 + BC 2 = 2AE 2 + 2iZF ; 
and, 

CD 2 + in 2 = 2Cl 2 + 2^^ 2 ; 

whence, by addition, recollecting that /l^ is equal to CE, 
and BE to ZLC 7 , we have, 




AB l + BO 2 + CD 2 + 2>yl 2 = ^C 7 ^ 2 + 4-fi# 2 ; 

but, 4CE 2 is equal to ylC 72 , and 4.DE 2 to 2/Zr 
(P. Vni., C.) : hence, 

AB 2 + BC 2 + CD 2 + DA 2 = AC 2 + BJf. 

That is, the sum of the squares of the sides of a paralklo- 
gram, is equal to the sum of the squares of its diagonals. 



110 



GEOMETRY. 



PROPOSITION XV. THEOREM. 

In any triangle, a line drawn parallel to the base divides 

the other sides proportionally. 

Let ABC be a triangle, and BE a line parallel to 
the base BC : then 



AB 



BB 



AE 



CK 



Draw EB and BC. Then, because 

the triangles AEB and BEB have their 

bases in the same line AB, and their 

vertices at the same point E, they will 

have a common altitude : hence, (P. VI., 

C.) 

AEB : BEB : : AB : BB. 



The triangles AEB and EBC, have their bases in the 
same line A C, and their vertices at the same point B ; 
they have, therefore, a common altitude ; hence, 




AEB : EBG 



AE : EC. 



But the triangles BEB and EBC have a common base 
BE, and their vertices in the line B C, parallel to BE ; 
they are, therefore, equal : hence, the two preceding propor- 
tions have a couplet in each equal ; and consequently, the 
remaining terms are proportional (B. II., P. IV.), hence, 



AB 



BB 



AE 



EC 



which teas to be proved. 



Cor. 1. We have, by composition (B. II., P. VI.), 
AB + BB : AB : : AE + EC : AE \ 



BOOK IV. 



Ill 



or, AB : AB : : AG : AE ; 

and, in like manner, 

AB : DB : : AG : EG, 

Gor. 2. If any number of parallels be drawn cutting two 
lines, they will divide the lines proportionally. 

For, let be the point where AB 
and GD meet. In the triangle OFF, 
the line A G being parallel to the base 
EF, we shall have, 

OE : AE : : OF : GF. 
In the triangle OGH, we shall have, 

OE i EG : : OE : FH ; 
hence (B. II, P. IV., C), 

AE : EG : : GF : FH. 
In like manner, 

EG : GB : : FH HD ; 

and so on. 




PEOPOSITION XVI. THEOEEM. 

If a line divides two sides of a triangle proportionally, it 
will be parallel to the third side. 

Let ABG be a triangle, and let BE 
divide AB and AG, so that 

AD : BB : : AE : EG ; 

then will BE be parallel to BG. 

Draw BG and EB. Then the tri- 




112 



GEOMETRY. 




angles ABE and BEB will have a common altitude ; and 
consequently, we shall have, 

ABE : BEB : : AB : BB. A 

The triangles ABE and EBG have also 
a common altitude ; and consequently, we 
shall have, 

ABE : EBG :: AE : EG; 

but, by hypothesis, 

AB i BB i: AE : EG ; 

hence (B. II., P. IV.), 

ABE : BEB : : ABE : EBG. 

The antecedents of this proportion being equal, the con- 
sequents will be equal ; that is, the triangles BEB and 
EBG are equal. But these triangles have a common base 
BE : hence, their altitudes are equal (P. VI., C.) ; that is, 
the points B and (7, of the line BC, are equally distant 
from BE, or BE prolonged : hence, BG and BE are 
parallel (B. I., P. XXX., C.) ; which icas to be proved. 



PROPOSITION XVII. THEOEEM. 

The line which bisects the vertical angle of a triangle, 
divides the base into segments proportional to the adja- 
cent sides. 

Let AB bisect the vertical angle A of the triangle 
BAG : then will the segments BB and BG be propor- 
tional to the adjacent sides BA and GA. 

From (7, draw CE parallel to BA, and produce it 



BOOK IV. 



113 



until it meets BA prolonged, at E Then, because CE 
and BA are parallel, the angles BAB and AEC are 
equal (B. I, P. XX., C. 3) ; the 
angles BAC and ACE are 
also equal (B. I, P. XX., C. 2). 
But, BAB and BAC are 
equa], by hypothesis ; consequent- 
ly, A EC and A CE are equal : 
hence, the triangle ACE is 
isosceles, AE being equal to 
AC. 

In the triangle BEC, the line AB is parallel to the 
base EC : hence (P. XV.), 

BA : AE : : BB : BC ; 

or, substituting AC for its equal AE, 

BA : AC : : BB : BC ; 

which was to be proved. 




PEOPOSITIO^ XVIII. THEOEEM. 
. Triangles xohich are mutually equiangular, are similar. 

Let the triangles ABC and BEE have the angle A 
equal to the angle B, the angle B to the angle E, and 
the angle C to the angle E : then will they be similar. 

For, place the triangle 
BEE upon the triangle 
ABC, so that the anjrle 
E shall coincide with the 
angle B then will the 

point F fall at some B* ^H C £ F 

point IT, of BC; the point 7> at some point G, of BA ; 

8 





\ 



114: GEOMETRY. 

the side DF will take the position GH, and BGH will 
be equal to EDF. 

Since the angle BUG 
is equal to BCA, GS 
will be parallel to AG 
(B. L, P. XIX., C. 2) ; 
and consequently, we shall 
have (P. XV.), 

BA : BG : : BC : BH 




H C 




or, since BG is equal to ED, and BH to EF, 
BA : ED : : BG : EF 

In like manner, it may be shown that 

BG i EF : : GA : FD ; 

and also, 

GA x FD : : AB : DE ; 

hence, the sides about the equal angles, taken in the same 
order, are proportional ; and consequently, the triangles are 
similar (D. 1) ; which loas to be proved. 

Gor. If two triangles have two angles in one, equal to 
two angles in the other, each to each, they will be similar 
(B. L, P. XXV., C. 2). 



PKOPOSITION XIX. THEOREM. 

Triangles which have their corresponding sides proportional, 

are similar. 

In the triangles ABG and DEF, let the corresponding 
^ides be proportional ; that is, let 



BOOK iv. ns 

AB : DE : : _£<7 : _£F : CA FD ; 

then will the triangles be similar. 

For, on BA lay off J? G equal to .ED; on BG tey 

off i?.^ equal to EF, 
and draw £i?: Then, 

because BG is equal to 

BE, and i?j^ to EF, 
we have, 

BA : BG : 





E F 



hence, G^ is parallel to A C (P. XVI.) ; and consequently, 
the triangles BAG and BGH are equiangular, and there- 
fore similar : hence, 

BG : BH : : CA : HG. 
But, by hypothesis, 

BG : EF : : (7J. : FD ; 

hence (B. H, P. IV., C), we have, 

BIT : EF : : JIG : FD. 

But, BII is equal to EF ; hence, .27# is equal to FD, 
The triangles BUG and .##£> have, therefore, their sides 
equal, each to each, and consequently, they are equal in all 
their parts. Now, it has just been shown that BUG and 
BGA are similar: hence, EFD and BGA are also simi- 
lar ; which was to be proved. 

Scholium. In order that polygons may be similar, they 
must fulfill two conditions: they must be mutually equian- 
gular, and the corresponding sides must he p>roportional. In 
the case of triangles, either of these conditions involves the 
other, which is not true of any other species of polygons. 





116 GEOMETRY. 

PROPOSITION XX. THEOREM. 

Triangles which have an angle in each equal, and the in* 
eluding sides proportional, are similar. 

In the triangles ABG and DEF, let the angle B be 
2qual to the angle E ; and suppose that 

BA : ED : : BO : EF ; 

then will the triangles "be similar. 

For, place the angle E 
upon its equal B ; F 
will fall, at some point of 

BG, as E; D will fall ^ 

at some point of BA, as B H C E F 

G ; DF will take the position GE, and the triangle 

DEF will coincide with GBE, and consequently, will be 

equal to it. 

But, from the assumed proportion, and because BG is 
equal to ED, and BEE to EF we have, 

BA : BG : : BG : BH -, 

hence, GH is parallel to AG ; and consequently, BAG 
and BGH are mutually equiangular, and therefore similar. But, 
EDF is equal to BGH: hence it is also similar to BAG', which 
v:as to he proved. 

PROPOSITION XXI. THEOREM. 

Triangles which have their sides parallel, each to each, or 
perpendicular, each to each, are similar. 

1°. Let the triangles ABG and DEF have the side 
AB parallel to DE, BG to EF, and CA to FD : 
then will they be similar. 



BOOK IV. 



117 



and B C 
(B. L, P. 





For, since the side AB is parallel to BE, 
to EF, the angle B is equal to the angle E 
XXIV.) ; in like manner, 
the angle C is equal to 
the angle F, and the an- 
gle A to the angle B ; 
the triangles are, therefore, 
mutually equiangular, and B ~~C 

consequently, are similar (P. XVIH) ; ichich icas to he 
proved. 

2°. Let the triangles ABC and BEF have the side 
AB perpendicular to BE, BG to EF, and CA to 
FB : then will they be similar. 

For, prolong the sides of the tri- 
angle BEF till they meet the sides 
of the triangle ABC. The sum of 
the interior angles of the quadrilateral 
BIEG is equal to four right angles 
(B.' L, P. XXVI.) ; but/ the angles 
FIB and EGB are each right 

angles, by hypothesis ; hence, the sum of the angles IEG 
IBG is equal to two right angles ; the sum of the angles 
IEG and BEF is equal to two right angles, because they 
are adjacent ; and since things which are equal to the same 
thing are equal to each other, the sum of the angles IEG 
^and IBG is equal to the sum of the angles IEG and BEF\ 
or, taking away the common part IEG, we have the angle 
IBG equal to the angle BEF. In like manner, the angle 
GCII may be proved equal to the angle FEB, and the 
angle MAI to the angle EBF ; the triangles ABC and 
BEF are, therefore, mutually equiangular, and consequently 
similar ; xoliich teas ■ to be x>roved. 

Cor. 1. In the first case, the parallel sides are homolo- 




118 



GEOMETRY. 



gous ; in the second case, the perpendicular sides are homo- 
logous. . 

Cor. 2. The homologous angles are those included by 
sides respectively parallel or perpendicular to each other. 

Scholium. When two triangles have their sides perpen- 
dicular, each to each, they may have a different relative 
position from that shown in the figure. But we can always 
construct a triangle within the triangle ABC, whose sides 
shall be parallel to those of the other triangle, and then the 
demonstration will be the same as above. 



PROPOSITION XXII. THEOEEM. 

If a line he drawn parallel to the base of a triangle, and 
lines be drawn from the vertex of the triangle to points 
of the base, these lines will divide the base and the par- 
allel proportionally. 

Let ABC be a triangle, BC its base, A its vertex, 
DE parallel to BC, and AF, AG, AH, lines drawn 
from A to points of the base : then will 

DI : BF : : IK : FG : : KL : GH : : LE : JSG. 



For, the triangles AID and 
AFB, being similar (P. XXL), we 
have, 

AI : AF : : DI : BF ; 

and, the triangles AIK and AFG, 
being similar, we have, 

AI : AF : : IK 




hence, (B. IL, P. IV.), we have, 



BOOK IV. 



119 



BI : BF : : IK : FG. 

In like manner, 

IK : FG : : XL : GH, 



and, 



KL : GH : : LE : HO ; 



hence (B. EL, P. IV.), 

DI : BF : : IK : FG : : KL : GH : : LE : HQ ; 

which was to be proved. 

Cor. If i?(7 is divided into equal parts at F, G, and 
H, then will BE be divided into equal parts, at I, K, 
and L. 



PROPOSITION XXIII. THEOREM. 

If, in a right-angled triangle, a perpendicular be drawn from 
' the vertex of the right angle to the hypothenuse : 

1°. The triangles on each side of the 2^?pendicidar will be 
similar to the given triangle, and to each other : 

2°. Each side about the right angle loill be a mean propor- 
tional between the hypothenuse and the adjacent segment : 

3°. The perpendicular will be a mean proportional between 
the two segments of the hypothenuse. 

1°. Let ABC be a right-angled triangle, A the vertex 
of the right angle, BC the hypo- 
thenuse, and AD perpendicular to 
BC : then will ABB and ABC 
be similar to ABC, and conse- 
quently, similar to each other. 

The triangles ABB and ABC 




have the angle B 



common, and the angles ABB and 



120 



GEOMETRY. 




BA G equal, because both are right angles ; they are, there- 
fore, similar (P. XVHL, C). In like manner, it may be 
shown that the triangles ABC and ABC are similar ; 
and since ABB and ABC are both similar to ABC, 
they are similar to each other ; which was to he proved. 

2°. AB will be- a mean pro- 
portional between B C and BB ; 
and AC will be a mean propor- 
tional between CB and CB. 

For, the triangles ABB and 
BAG being similar, their homo- 
logous sides are proportional : hence, 

BC : AB : : AB : BB. 

In like manner, 

BC : AC : : AC : BC ; 
which was to be proved. 

3°. AB will be a mean proportional between BB and 
BC. For, the triangles ABB and ABC being similar, 
their homologous sides are proportional ; hence, 

BB : AB : : AB • BC ; 

which was to be proved. 

Cor. 1. From the proportions, 

BC : AB : : AB : BB, 
BC -.AC : : AC : BC, 
we have (B. II., P. I.), 



and, 



and, 



AB" = BC x BB, 



AC 1 = BC x BC ; 



BOOK IV. 



121 



whence, by addition, 



AB* + AC - BC (BD + DC) ; 



or. 



AB* + AC = BG' ; 



as was shown in Proposition XI. 



Cor. 2. If from any point A, in a semi-circumference 
BAG, chords be drawn to the 
extremities B and G of the diam- 
eter BG, and a perpendicular AD 
be drawn to the diameter : then 
will ABC be a right-angled tri- 
angle, right-angled at A ; and from what was proved above, 
each chord will be a mean proportional between the diameter 
and the adjacent segment ; and, the perpendicidar will be a 
mean proportional between the segments of the diameter. 




PROPOSITION XXIV. THEOREM. 



Triangles which have an angle in each equal, are to each 
other as the rectangles of the including sides. 

Let the triangles GJJK and ABC have the angles G 
and A equal : then will they be to each other as the 
rectangles of the sides about these angles. 

For, lay off AD equal 



to Gil, AE to GK, and 

draw DE ; then will the 
triangles ADE and GJIJC 
be equal in all their parts. 
Draw EB. 



G 



J I 




K 




122 



GEOMETRY. 



The triangles ABE and ABE have their bases in the 

same line AB, and a common vertex E ; therefore, they 

have the same altitude, and consequently, are to each other 
as their bases ; that is, 



ABE 



ABE 



AB 



AB. 





The triangles ABE and q 

ABC, have their bases in 

the same line AC, and a H 

common vertex B ; hence, ^K 

ABE : ABC : : AE : AC; 

multiplying these proportions, term by term, and omitting 
the common factor ABE (B. II., P. VII.), we have, 

ABE : ABC : : AB x AE : AB x AC ; 

substituting for ABE, its equal, GEE, and for AB x AE, 
its equal, GH X GK, we have, ' 

GHK : ABC : : GH x GK : AB x AC ; 

w/b'cA was to he proved. 

Cor. If ^.£>i? and ABC are similar, the angles B 
and i? being homologous, BE will be parallel to BC, 
and we shall have, 

AB : AB : : AE : AC ; 
hence (B. II., P. IV.), we have, 
ABE : ^i?i? : : ABE : ABC ; 

that is ABE is a mean proportional be- 
tween ABE and ^JSC. 




BOOK IV. 



123 



PKOPOSITIOX XXV. THEOKEM. 

Similar triangles are to each other as the squares of their 

homologous sides. 

Let the triangles ABC and BEF be similar, the angle 
A being equal to the angle Z>, B to E, and G to F. 
then will the triangles be to each other as the squares of 
any two homologous sides. 

Because the angles A and D are equal, we have (P. 
XXIY.), 

ABC : DEF :: AB x AC : DE x BF ; 



and, because the triangles 
are similar, we have, 

AB : BE n AG : BF; 

multiplying the terms of ' / — 
this proportion by the cor- 
responding terms of the proportion, 




we have (B. II, P. XII.), 




AC : BF : : AC : BF, 



AB x AC : BE x BF : : AC 2 : BF 2 ; 

combining this, with the first proportion (B. II., P. IV".), 
we have, 

ABC : BEF : : AC 2 : BF\ 

In like manner, it may be shown that the triangles are 

to each other as the squares of AB and BE, or of BC 
and EF ; which icas to be proved. 



124: GEOMETRY. 



PROPOSITION XXVI. THEOREM. 

Similar polygons may be divided into the same number of 
triangles, similar, each to each, and similarly placed. 

Let ABCDE and FGHIK be two similar polygons, 
the angle A being equal to the angle F, B to G, C to 
IT, and so on : then can they be divided into the same 
number of similar triangles, similarly placed. 

For, from A draw 

the diagonals AC , C 

AD, and from F, B r — ^X\ r ^-— -A 

homologous with A , \ / \- I / \ T 

° . .1/ ->D1/^ ~->l 

draw the diagonals A\^" y/ *\. ^^ 

i^ i^T, to the ver- ^^^X K 

E 

tices ^" and I, hom- 
ologous with (7 and D. 

Because the polygons are similar, the triangles AJBO and 
FGH have the angles B and 6r equal, and the sides 
about these angles proportional ; they are, therefore, similar 
(P. XX.). Since these triangles are similar, we have the 
angle AGB equal to FUG, and the sides AG and FH, 
proportional to BC and GH, or to CD and HI. The 
angle BCD being equal to the angle GDI, if we take 
from the first the angle ACB, and from the second the 
equal angle FUG, we shall have the angle A CD equal 
to the angle Fill : hence, the triangles A CD and FHI 
have an angle in each equal, and the including sides propor- 
tional ; they are therefore similar 

In like manner, it may be shown that ADE and FIA 
are similar ; which icas to be proved. 

Cor. 1. The corresponding triangles in the two polygons 
are homologous triangles, and the corresponding diagonals are 
homologous diagonals. 



BOOK IV. 125 

Cor. 2. Any two homologous triangles are like parts of 
the polygons to which they belong. 

For, ABC and FGH being similar, we have, 

ABC : FGH : : J^" 2 : J^ 2 ; 

and, for a hke reason, 



ACJD : FHI : : AC 2 : FH 2 ; 
whence, 

ABC : FGH :: ACD : FHI ; 

and, in like manner, 

^<7Z> : FHI : : ABE : IFF. 

Cor. 3. If two polygons are made up of similar triangles 
similarly placed, the polygons themselves will be similar. 



PEOPOSITIOX XXVII. THEOEE3I. 

The jwrimeters of similar polygons are to each other as any 
two homologous sides ; and the polygons are to each 
other as the squares of any two homologous sides. 

1°. Let ABCBE and FGHIK be similar polygons: 
then will their perimeters be to each other as any two 
homologous sides. 

For, any two homo- 
logous sides, as AB 
and FG, are like parts 
of the perimeters to 
which they belong : 
hence (B. II., P. IX.), 
the perimeters of the 

polygons are to each other as AB to FG, or as any 
other two homologous sides ; which teas to be proved. 




126 



GEOMETRY. 




2°. The polygons will be to each, other as the squares 
of any two homologous sides. 

For, let the poly- 
gons be divided into 
homologous triangles 
(P. XXVI., C. 1) ; 
then, because the 
homologous triangles 
ABC and FGJEC are 

like parts of the polygons to which they belong, the poly- 
gons will be to each other as these triangles ; but these 
triangles, being similar, are to each other as the squares of 
AB and FG : hence, the polygons are to each other as 
the squares of AB and FG, or as the squares of any 
other two homologous sides ; which was to be proved. 

Cor. 1. Perimeters of similar polygons are to each other 
as their homologous diagonals, or as any other homologous 
lines ; and the polygons are to each other as the squares of 
their homologous diagonals, or as the squares of any other 
homologous lines. 

Cor. 2. If the three sides of a right-angled triangle be 
made homologous sides of three similar polygons, these poly- 
gons will be to each other as the squares of the sides of 
the triangle. But the square of the hypothenuse is equal 
to the sum of the squares of the other sides, and conse- 
quently, the polygon on the hypothenuse will be equal to 
the sum of the polygons on the other sides. 



PROPOSITION XXVIII. THEOREM. 

If two chords intersect in a circle, their segments will be 

reciprocally proportional. 

Let the chords AB and CD intersect at : then 




BOOK IV. !27 

will their segments be reciprocally proportional ; that is, one 
segment of the first will be to one segment of the second, 
as the remaining segment of the second is to the remaining 
segment of the first. 

For, draw CA and BB. Then 
will the angles OBB and OAC be 
equal, because each is measured by half 
of the arc CB (B. UT., P. XVIII.) . 
The angles OBB and OCA, will also 
be equal, because each is measured by 

half of the arc AB : hence, the triangles OBB and CA 
are similar (P. XVIII., C), and consequently, their homolo- 
gous sides are proportional : hence, 

BO : AO : : OB : OC ; 
which was to be proved. 

Cor. From the above proportion, we have, 

BO x OC = AO x OB ; 

that is, the rectangle of the segments of one chord is equal 
to the rectangle of the segments of the other. 



PROPOSITION XXIX. THEOREM. 

If from a point without a circle, two secants be drawn ter- 
minating in the concave arc, they will be reciprocally 
proportional to their external segments. 

Let OB and OC be two secants terminating in the 
concave arc of the circle BCB : then will 

OB : OC : : OB : OA. 



128 



GEOMETRY. 



For, draw AC and BB. The triangles OJDB and 
OAC have the angle common, and the angles OBB 
and OCA equal, because each is measured 
by half of the arc AD : hence, they are 
similar, and consequently, their homologous 
sides are proportional ; whence, 



OB 



oc 



OB 



OA ; 



which was to be proved. 




Cor. From the above proportion, we 

have, 

OB x OA = OC x OB ; 

that is, the rectangles of each secant and its external seg- 
ment are equal. 



PROPOSITION XXX. THEOREM. 

If from a point loithout a circle, a tangent and a secant 
be drawn, the secant terminating in the concave arc, the 
tangent will be a mean proportional between the secant 
and its external segment. 

Let ABC be a circle, OC a secant, and OA a tan- 
gent : then will 

OC : OA : : OA : OB. 



For, draw AB and AC. The tri- 
angles OAB and OAC will have the 
angle common, and the angles OAB 
and A CB equal, because each is mea- 
sured by half of the arc AB (B. III., 
P. XVIIL, P. XXI.) ; the triangles are 
therefore similar, and consequently, their 




BOOK IV. 129 

homologous sides are proportional : hence, 

00 : OA : : OA : OB ; 
which icas to he pi*oved. 

Cor. From the above proportion, we have, 

AO 2 = OC x OD ; 

that is, the square of the tangent is equal to the rectangle 
of the secant and its external segment. 



PEACTICAL APPLICATIONS. 



PROBLEM I. 

To divide a given line into parts proportional to given lines, 

also into equal parts. 

1°. Let AB be a given line, and let it be required to 
divide it into parts proportional to the lines B, Q and B. 

From one extremity A, 
draw the indefinite line AG, 
making any angle with AB ; 
lay off A C equal to B, OB 
equal to Q, and BE equal 
to B ; draw EB, and 
from the points C and B, 
draw CI and BF parallel to EB : then will AT, IF, 
and EB, be proportional to B, Q, and B (P XV., C. 2). 

9 




130 



GEOMETRY. 



2° Let AH be a given line, and let it be required to 
divide it into any number of equal parts, say five. 

From one extremity 

A, draw the indefinite ^ C 

line A G ; take A I equal 

to anv convenient line, 

and lay off IE, KZ, 

ZM, and MB, each 

equal to AI. Draw 

BE, and from I, K, Z, and M, draw the 4ines IC, 

KB, ZE, and ME, parallel to BE : then will AS be 

divided into equal parts at C, B, E, and F (P. XV., 

C. 2). 

PROBLEM II. 




To construct a fourth proportional to three given lines. 

Let A, B, and C, be 
the given lines. Draw 
BE and BE, making 
any convenient angle with 
each other. Lay off BA 
equal to A, BB equal 
to B, and BC equal 

to C ; draw AC, and from B draw BX parallel to 
AC : then will BX be the fourth proportional required. 
For (P. XV., C), we have, 




BA : BB : : BC : X> X ; 



or, 



^ : B 



C : £>X 



6 7 or. If BC is made equal to BB t BX will be 
tltird proportional to BA and BB, or to ^4 and B, 



BOOK IV. 



131 



PEOBEEM in. 



To construct a mean proportional between two given lines. 

Let A and B bo the given 
lines. On an indefinite line, lay off 
BE equal to A, and EF equal 
to B ; on DF as a diameter de- 
scribe the semi-circle BGF, and 
draw EG perpendicular to BE: 
then will EG be the mean proportional required 

For (P. XXm, C. 2), we have, 




or, 



BE : EG : : EG : EF ; 
^ : EG : : EG : B. 



PEOBLEM IY. 



To divide a given line into two such parts^ that the greater 
part shall be a mean proportional between the whole line 
and the other part. 

Let AB be the given line. 

At the extremity _Z>, draw 
BC perpendicular to AB, and 
make it equal to half of AB. 
With C as a centre, and CB 
as a radius, describe the are 
BBE ; draw A C\ and produce 
it till it terminates in the concave arc at E ; with A as 
centre and AB as radius, describe the arc BF : then 
will AF be the greater part required. 




132 



GEOMETRY. 



For, AB being perpendicular to CB at B, is tan* 
gent to the arc BBE : hence 
(P. XXX.), 



AE 



AB 



AB 



AB-, 



and, by division (B. II., P. VI.), 




AE - AB : AB : : AB - AB : AB. 



But, BE is equal to twice CB, or to AB : hence, 
AE — AB is equal to AB, or to AE; and AB — AB 
is equal to AB — AE, or to FB : hence, by substitution, 

fa 

AE : AB : : FB : AE ; 

and, by inversion (B. II., P. V.), 

AB : AE : : AE : FB. 

JSchoUum. When a line is divided so that the greater 
segment is a mean proportional between the w^hole line and 
the less segment, it is said to be divided in extreme and 
mean ratio. 

Since AB and BE are equal, the line AE is divided 
in extreme and mean ratio at B ; for we have, from the 
first of the above proportions, by substitution, 



AE : BE : : BE : AB. 



BOOK IV. 



133 



PROBLEM V. 

* 
Through a given point, in a given angle, to draw a line 

so that the segments between the point and the sides of 

the angle shall be equal. 

Let BCD be the given angle, and A the given point, 

Through A, draw AE parallel to 
DC ; lay off EF equal to CE, and 
draw FAD : then will AF and AD, K/ 

be the segments required. p. 

For (P. XV.), we have, 

FA : AD : : FE : EC ; B ' 
but, FE is equal to EC ; hence, FA is equal to AD, 





PROBLEM VI. 
To construct a triangle equal to a given polygon. 

Let ABCDE be the given polygon. 

Draw CA ; produce EA, and 
draw BG parallel to CA ; draw 
the lino CG. Then the triangles 
BAC and GAC have the com- 
mon base AC, and because their 
vertices B and G lie in the 
• same line BG parallel to the base, their altitudes are equal, 
and consequently, the triangles are equal : hence, the polygon 
GCDE is equal to the polygon ABCDE. 

Again, draw CE ; produce AE and draw DF parallel 
to CE ; draw also CF ; then will the triangles FCE 
and DCE be equal: hence, the triangle GCF is equal 
to the polygon GCDE, and consequently, to the given 
polygon. In like manner, a triangle may be constructed 
equal to any other given polygon. 



134: 



GEOMETRY. 



PEOBLEM VIE 

To construct a square equal to a given triangle. 

Let ABG be the given triangle, AD its altitude, and 
MG its base. 

Construct a mean pro- 
portional between AD 
and half of BG (Prob. 
HI). Let XT be that 
mean proportional, and on 
it, as a side, construct a 

square : then will this be the square required. For, from 
the construction, 




C X 




XT = \BG X AD = area AD G. 

■ Scholium. By means of Problems VI. and VIL, a square 
may be constructed equal to any given polygon. 



PEOBLEM VIII. 

On a given line, to co?istruct a polygon similar to a given 

polygon. 

Let FG be the given line, and ABGDE the given 
polygon. Draw AG and AD. 

At F, construct 
the angle GFH equal 
to BAG, and at G 
the angle FGII equal 
to ABG ; then will 
FGII be similar to 
ABG (P. XVIIL, C.) 




BOOK IV. 



135 



In like manner, construct the triangle FHI similar to 
A CD, and FIK similar to ADE '; then will the polygon 
FGHIK be similar to the polygon ABCDE (P. XXVI., 
P.). 




PROBLEM LX. 

To construct a square equal to the sum of two given 
squares, also a square equal to the difference of two 
given squares. 

1°. Let A and B be the sides of the given squares, 
and let A be the greater. 

Construct a ri^ht angle 
CDE ; make DE equal 
to A, and DC equal to 
B ; 'draw CE, and on it 
construct a square : this square will be equal to the sum 
of the given squares (P. XL). 

2°. Construct a right angle CDE. 

Lay off DC equal to B ; with C 
as a centre, and CE, equal to A, as 
a radius, describe an arc cutting DE at 
E ; draw CE, and on DE construct 
a square : this square will be equal to 
the difference of the given squares (P. XL, C. 1). 

Scholium. By means of Probs. VL, VIL, VIIL, and IX. 

a polygon may be constructed similar to either of two given 
similar polygons, and equal to their sum, or to their difference. 




BO OK Y. 

EEGULAE POLYGONS. — AREA OF TE^ CIECLE. 

DEFINITION. 

1. A Regular Polygon is a polygon which is "both 
equilateral and equiangular. 



PROPOSITION I. THEOREM. 

Regular polygons of the same member of sides are similar. 

Let ABCDJEF* and abedef be regular polygons of the 
same number of sides : then will they be similar. 

For, the corresponding 
angles in each are equal, 
because any angle in 
either polygon is equal 
to twice as many right 

angles as the polygon ^ ■£ 

has sides, less four, di- 
vided by the number of angles (B. I., P. XXVI., C. 4) ; and 
further, the corresponding sides are proportional, because all 
the sides of either polygon are equal (D. 1) : hence, the 
polygons are similar (B. IY., D. 1) ; loJiich was to be proved. 







BOOK V. 



137 



PKOPOSITIOX II. THEOEEM. 



The circumference of a circle may be circumscribed about any 
regular polygon ; a circle may also be inscribed in it. 

1°. Let AJBCF be a regular polygon : then can the 
circumference of a circle be circumscribed about it. 

For, through three consecutive ver- 
tices A, B, C, describe the circum- 
ference of a circle (B. ITT., Problem 
XIII., S.). Its centre Trill lie 
on PO, drawn perpendicular to BC, 
at its middle point P; draw OA 
and OD. 

Let the quadrilateral OP CD be 
turned about the line OP, until PC 

falls on PB ; then, because the angle C is equal to _Z>, 
the side CD will take the direction BA ; and because CD 
is equal to BA, the vertex D, will fall upon the vertex 
A ; and consequently, the line OD will coincide with OA, 
and is, therefore, equal to it : hence, the circumference which 
passes through A, B, and C, will pass through D. In 
like manner, it may be shown that it will pass through all 
of the other vertices : hence, it is circumscribed about the 
polygon ; xcldch was to be proved. 




4. . 



A circle may be inscribed in the polygon. 
For, the sides AB, BC, &c, being equal chords of 
the circumscribed circle, are equidistant from the centre 
hence, if a circle be described from as a centre, with 
OP as a radius, it will be tangent to all of the 6ides or 
the polygon, and consequently, will be inscribed in it; which 
icas to be proved. 



138 



GEOMETRY. 



Scholium. If the circumference of a circle be divided 
into equal arcs, the chords of these arcs will be sides of a 
regular inscribed polygon. 

For, the sides are equal, because they are chords of equal 
arcs, and the angles are equal, because they are measured by 
halves of equal arcs. 

If the vertices A, JB, C, &c, 
of a regular inscribed polygon be 
joined with the centre 0, the tri- 
angles thus formed will be equal, 
because their sides are equal, each 
to each : hence, all of the angles 
about the point are equal to 
each other. 




DEFINITIONS. 

1. The Centre of a Regular Polygon, is the common 
centre of the circumscribed and inscribed circles. 

2. The Angle at the Centre, is the angle formed by 
drawing lines from the centre to the extremities of either 
side. 

The angle at the centre is equal to four right angles 
divided by the number of sides of the polygon. 



3. The Apothem, is the distance from the centre to 
either side. 

The apotl em is equal to the radius of the inscribed 
circle. 



BOOK V. 



139 



PEOPOSITION in. PEOBLEM. 

To inscribe a square in a given circle. 

Let ABCD be the given cir- 
cle. Draw any two diameters A C 
and BD perpendicular to each 
other ; they will divide the circum- 
ference into, four equal arcs (B. HI., 
P. XVn., S.). Draw the chords 
AB, B C, CD, and DA : then 
will the figure ABCD be the 
square required (P. H., S.). 

Scholium. The radius is to the side of the inscribed 
square as 1 is to ypl. 




PROPOSITION IV. 



THEOEEM. 



If a regular hexagon be inscribed in a circle, any side will 
be equal to the radius of the circle. 

Let ABD be a circle, and ABCDJEH a regular in- 
scribed hexagon : then will any side, as AB, be equal to 
the radius of the circle. 

Draw the radii OA and OB. 
Then will the angle A OB be 
equal to one-sixth of four right 
angles, or to two-thirds of one 
right angle, because it is an an- 
gle at the centre (P. II., D. 2). 
The sum of the two angles OAB 
and OBA is, consequently, equal 

to four-thirds of a right angle (B. L, P. XXV., C. 1) ; but, 
the angles OAB and OBA are equal, because the opposite 
sides OB and OA are equal : hence, each is equal to 




140 



GEOMETRY. 



two-thirds of a right angle. The three angles of the triangle 
A OB are therefore, equal, and consequently, the triangle is 
equilateral : hence, AB is equal to OA ; which teas to be 
proved. 

PEOPOSITIOjST Y. PEOBLE3I. 
To inscribe a regular hexagon in a given circle. 

Let ABE be a circle, and its centre. 

Beginning at any point of 
the circumference, as A, ap- 
ply the radius OA six times 
as a chord ; then will 
ABCDEF he the hexagon 
required (P. IV.). 

Cor. 1. If the alternate 

vertices of the regular hexagon 

be joined by the lines AC, 

CE, and EA, the inscribed 

triangle ACE will be equilateral " (P. II., S.). 

Cor. 2. If we draw the radii OA and OC, the figure 
AOCB will be a rhombus, because its sides are equal: 
hence (B. IV., P. XIV., C), we have, 




AB 2 + BC 2 -f- OA 2 + OC' = AC + OB'; 

or, taking away from the first member the quantity OA , 
and from the second its equal OB 2 , and reducing, we have 



3 0A* = AC-, 



whence (B. II., P II.) , 



AC 2 : OA 2 : : 3 : 1; 



BOOK V. 



141 



or (B. n., P. xn., C. 2), 

A C : A : : y/i : 1 ; 

that is, the side of an inscribed equilateral triangle is to the 
radius, as the square root of 3 is to 1. 



PROPOSITION VI. 



THEOREM. 



If the radius of a circle be divided in extreme and mean 
ratio, the greater segment will be equal to one side of a 
regular inscribed decagon. 

Let ACG be a circle, OA its radius, and AB, equal to 
031, the greater segment of OA when divided in extreme 
and mean ratio : then will AB be equal to the side of a 
regular inscribed decagon. 

Draw OB and BM. We 
have, by hypothesis, 

AO : 031 : : 031 : AM; 

or, since AB is equal to 
031, we have, 

AO : AB : : AB : AM; 

hence, the triangles OAB 
and BA3L have the sides 
about their common angle 

BA3I, proportional ; tliey are, therefore, similar (li. IV., 
P. XX.). But, the triangle OAB is isosceles ; hence, BAM 
is also isosceles, and consequently, the side B3I is equal to 
AB. But, AB is equal to 031, by hypothesis : hence, 
B3I is equal to 031, and consequently, the angles MOB 




142 



GEOMETRY. 



and MBO are equal. The angle A21B being an exterior 
angle of the triangle OMB, is equal to the sum of the 
angles MOB and MBO, or 
to twice the angle MOB ; 
and because AMB is equal to 
OAB, and also to OBA, the 
sum of the angles OAB and 
OBA is equal to four times 
the angle A OB : hence, A OB 
is equal to one-fifth of two 
right angles, or to one-tenth of 
four right angles ; and conse- 
quently, the arc AB is equal 
to one-tenth of the circumfer- 
ence : hence, the chord AB is equal to the side of a 
regular inscribed decagon ; ichich was to be proved. 

Cor. 1. If AB be applied ten times as a chord, the 
resulting polygon will be a regular inscribed decagon. 

Cor. 2. If the vertices A, (7, E, G y and 7, of the 
alternate angles of the decagon be joined by straight lines, 
the resulting figure will be a regular inscribed pentagon. 




Scholium 1. If the arcs subtended by the sides of any 
regular inscribed polygon be bisected, and chords of the semi- 
arcs be drawn, the resulting figure will be a regular inscribed 
polygon of double the number of sides. 

Scholium 2. The area of any regular inscribed polygon 
is less than that of a regular inscribed polygon of double 
the number of sides, because a part is less than the whole 



BOOK V. 



U'S 



FKOPOSITION VH. PROBLEM. 

To circumscribe a polygon about a circle which shall be 
similar to a given regular inscribed polygon. 



Let TNQ be a circle, 
a regular inscribed polygon. 

At the middle points 
T, N, P, &c, of the arcs 
subtended by the sides of 
the inscribed polygon, draw 
tangents to the circle, and 
prolong them till they in- 
tersect ; • then will the re- 
sulting figure be the poly- 
gon required. 

1°. The side HG> be- 
ing parallel to BA, and 



its centre, and ABCDEF 




In 



HI to BC, the angle H is equal to the angle B. 
like manner, it may be shown that any other angle of the 
circumscribed polygon is equal to the corresponding angle of 
the inscribed polygon : hence, the circumscribed polygon is 
equiangular. 

2°. Draw the lines OG, OT, Oil, ON", and 01 Then, 
because the lines IIT and UN are tangent to the circle, 
Oil will bisect the angle NUT, and also the angle NOT 
(B. III., Prob. XIV., S.) ; consequently, it will pass through 
the middle point B of the arc NBT. In like manner, it- 
may be shown that the line drawn from the centre to the 
vertex of any other angle of the circumscribed polygon, will 
pass through the corresponding vertex of the inscribed poly- 
gon. 

The triangles OIIG and OIII have the angles OIIO 



1U 



GEOMETRY. 




and OSI equal, from what has just been shown ; the an- 
gles GOS and SOI equal, "because they are measured by 
the equal arcs AJ3 and 
BC, and the side OS 
common ; they are, there- 
fore, equal in all their 
parts : hence, GS is 
equal to SI In like 
manner, it may be shown 
that SI is equal to IS, 
IS to SI, and so on : 
hence, the circumscribed 
polygon is equilateral. 

The circumscribed poly- 
gon being both equiangular and equilateral, is regular ; and 
since it has the same number of sides as the inscribed poly- 
gon, it is similar to it. 

Cor. 1. If lines be drawn from the centre of a regular 
circumscribed polygon to its vertices, and the consecutive points 
in which they intersect the circumference be joined by ' 
chords, the resulting figure will be a regular inscribed 
polygon similar to the given polygon. • 

Cor. 2. The sum of the lines ST and SN is equal 
to the sum of ST and TG, or to SG ; that is, to one 
of the sides of the circumscribed polygon. 

Cor. 3. If at the vertices A, B, C, &c, of the in- 
scribed polygon, tangents be drawn to the circle and pro- 
longed till they meet the sides of the circumscribed polygon, 
the resulting figure will be a circumscribed polygon of double 
the number of sides. 



Cor. 4. The area of any regular circumscribed polygon 



I 



BOOK V. 



145 



is greater than that of a regular circumscribed polygon of 
double the number of sides, because the whole is greater 
than any of its parts. 

Scholium. By means of a circumscribed and inscribed 
square, we may construct, in succession, regular circumscribed! 
and inscribed polygons of 8, 16, 32, <fcc, sides. By means 
of the regular hexagon, we may, in like manner, construct 
regular polygons of 12, 24, 48, &c, sides. By means of the 
decagon, we may construct regular polygons of 20, 40, 80, 
&c., sides. 



PROPOSITION VIII. THEOREM. 

The area of a regular polygon is equal to half the product 

of its perimeter and apothem. 

Let GIIIK be a regular polygon, its centre, and 
OT its apothem, or the radius of the inscribed circle : 
then will the area of the polygon be equal to half the 
product of the perimeter and the apothem. 

For, draw lines from the centre 
to the vertices of the polygon. 
These lines will divide the polygon 
into triangles whose bases will be 
the sides of the polygon, and 
whose altitudes will be equal to 
the apothem. Now, the area of 
any triangle, as OJIG, is equal to 
half the product of the side JIG 
and the apothem : hence, the area 

of the polygon is equal to half the product of the perimeter 
and the apothem ; which was to be proved. 

10 




146 



GEOMETRY. 



PROPOSITION IX. THEOREM. 

The perimeters of similar regular polygons are to each 
other as the radii of their circumscribed or inscribed 
circles ; and their areas are to each- other as the squares 
of those radii. 

1°. Let ABO and KLM be similar regular polygons. 
Let OA and QK be the radii of their circumscribed, OB 
and QB be the radii of their inscribed circles : then will 
the perimeters of the polygons be to each other as OA is 
to QK, or as OB is to QB. 

For, the lines 
OA and QK are A, 

homologous lines 
of the polygons 
to which they be- 
long, as are also 
the lines OB and 

QB : hence, the 

perimeter of AB C 

is to the perimeter of KLM, as OA te to QK, or as 

OB is to QB (B. IV., P. XXVII., C. 1) ; which was to be 

proved. 

2°. The areas of the polygons will be to each other as 
OA 2 is to QK\ or as OB 2 is to QB 2 . 

For, OA being homologous with QK, and OB with 
QB, we have, the area of ABC is to the area of KLM 
as OT is to QK\ or as OB 2 is to QB 2 (B. IV., P 
XXVIL, C. 1) ; which was to be proved. 





BOOK V. 



147 



PROPOSITION X. THEOEEM. 

Two regular polygons of the same number of sides can be 
constructed, the one circumscribed about a circle and the 
other inscribed in it, which shall differ from each other 
by less than a?iy given surface. 

Let ABCE be a circle, O its centre, and Q the side 
of a square which is less than the given surface ; then can 
two similar regular polygons be constructed, the one circum- 
scribed about, and the other inscribed within the given circle 
which shall differ from each other by less than the square 
of g, and consequently, by less than the given surface. 

Inscribe a square in the 
given circle (P. III.), and by 
means of it, inscribe, in succes- 
sion, regular polygons of 8, 16, 
32, &c, sides (P. VH., S.), un- 
til one is found whose side is 
less than Q ; let AB be the 
side of such a polygon. 

Construct a similar circum- 
scribed polygon abode : then 
will these polygons differ from each other by less than the 
square of Q. 

For, from a and b, draw the lines a and b ; they 
will pass through the points A and B. Draw also OK 
to the point of contact K\ it will bisect AB at I and 
be perpendicular to it. Prolong AO to M 

Let P denote the circumscribed, and p the inscribed 
polygon ; then, because they are regular and similar, we 
shall have (P. IX.), 




148 



G E O IVI E T R Y. 



P : p : : OK or OA : 01 ; 



hence, by division (B, II., P. VI.), we have, 



P : P -p : : OA 



OA 



or 



or, 



AI\ 




P : P -p : : J. 2 : 

Multiplying the terms of the 
second couplet by 4 (B. II., P. 
VII), we have, 

P : P-^ : : 4<M 2 : 4 JT 2 ; 

whence (B. IV., P. VIIL, C), 

P : P-p : : AE 2 : AB\ 

But P is less than the square of AE (P. VII., C. 4) ; 
hence, P — p is less than the square of A£, and conse- 
quently, less than the square of Q, or than the given sur- 
face ; which was to be proved. 

Gov. 1. When the number of sides of the inscribed poly- 
gon is increased, the area of the polygon will be increased, 
and the area of the corresponding circumscribed polygon will 
be diminished (P. ' VII, c. 4) ; and each will constantly 
approach the circle, which is the limit of both. 

Cor. 2. When the number of sides of either polygon 
reaches its limit, which is infinity, each polygon will reach 
its limit, which is the circle: hence, under that supposition, 
the difference between the two polygons will be less than 
any assignable quantity, and may be denoted by zero* and 
either of the polygons will be represented by the circle. 

* Univ. Algebra, Arts. 72, 73. Bourdon, Art. 71. 



BOOK V 



149 



Scholium 1. The circle may be regarded as the l: .mit 
of the inscribed and circumscribed polygons ; that is. : X is 
a figure towards which the polygons may be made * ■ ap- 
proach nearer than any appreciable quantity, but b^ond 
which they cannot be made to pass. 

Scholium 2. The circle may, therefore, be regarded as a 
regular polygon of an infinite number of sides; and because 
of the principle, that whatever is true of a whole class, is 
true . of every individual of that class, we may affirm that 
ichatever is true of a regular polygon, having an infinite 
number of sides, is true also of the circle. 

Scholium 3. When the circle is regarded as a regular poly- 
gon, of an infinite number of sides, the circumference is to be 
regarded as its perimeter, and. the radius as its apothem. 

PROPOSITION XL PROBLEM. 

The area of a regular inscribed polygon, and that of a 
similar circumscribed polygon being given, to find the 
areas of the regular inscribed and circumscribed polygons 
having double the number of sides. 

Let AB be the side of the given inscribed, and EF 
that of the given circumscribed polygon. Let C be their 
common centre, AMB a portion of the circumference of 
the circle, and M the middle point of the arc AMB. 

Draw the chord AM, and 
at A and B draw the tangents 
AP and BQ ; then will AM 
be the side of the inscribed 
polygon, and BQ the side of 
the circumscribed polygon of 
double the number of sides (P. 
VII.). Draw CE, CB, CM, 
and CF. 




150 



GEOMETRY. 



Denote the area of the given inscribed polygon by p, 
the area of the given circumscribed polygon by P, and the 
areas of the inscribed and circumscribed polygons having 
double the number of sides, respectively by p' and P'. 

1°. The triangles CAB, CAM, 
and CEM, are like parts of the 
polygons to which they belong : 
hence, they are proportional to the 
polygons themselves. But CAM 
is a mean proportional between 
CAB and CEM (B. IV., P. 
XXIV., C.) ; consequently p' 
is a mean proportional between 
p and JP: hence, 




p' = y> x P. 



(i.) 



2°. Because the triangles CPM and CPE have the 
common altitude CM, they are to each other as their 
bases : hence, 

CPM : CPE : : PM : PE ; 

and because CP bisects the angle ACM, we have (B. IV., 
P. XVII.), 

PM : PE : : CM : CE : : CB ; CA ; 



hence (B. II., P. IV.), 
CPM : CPE 



CB 



CA or CM. 



But, the triangles CAB and CA31 have the common 
altitude AB ; they are therefore, to each other as their 
bases : hence, 

CAB : CAM : : CD : C M ; 



BOOK V. 151 

or, because CAD and CAM are to each other as the 
polygons to which they belong, 

p : p' : : CD : CM ; 
hence (B. II., P. IV.), we have, 

CPM : CPE : : p : p\ 
and, by composition, 

CPM : CPM + CPU or CME : : p : jp+i?' ; 
hence (B. H, P. VII.), 

2 OPJjr or 0JQM : CME : : 2p : p + p'. 

Bat, CMP A and tfjlftf are like parts of P' and P, 

hence, 

P : P : : 2p : p + p r ; 



or, 



j* = 2 ^ xP f (2.) 

p+p' v ' 

Scholium. By means of Equation ( 1 ), we can find p\ 
and then, by means of Equation (2), we can find P '. 



PROPOSITION XII. PROBLEM. 
To find the approximate area of a circle whose radius is 1. 

The area of an inscribed square is equal to twice the 
square of the radius, or 2 (P. III., S.), and the area of a 
circumscribed square is 4. Making p equal to 2, and 
V equal to 4, we have, from Equations ( 1 ) and ( 2 ) of 
Proposition XI., 

p' = ys — 2.8284271 . . . inscribed octagon ; 
P' = r — 3.3137085 . . . circumscribed octagon. 



2 +^8 



152 



GEOMETRY. 



Making p equal to 2.8284271, and P equal to 3.3137085, 
we have, from the same equations, 

p' = 3.061/4674 . . . inscribed polygon of 16 sides. 
P' = 3.1825979 . . . circumscribed polygon of 16 sides. 

By a continued application of these equations, we find 
the areas indicated in the following 





TABLE. 




Number op Sides. 


INSCRIBED POLYGONS. 


Circumscribed Polygons, 


4 . 


2.0000000 


4.0000000 


8 


2.8284271 


3.3137085 


16 


3.0614674 


3.1825979 


32 


3.1214451 


3.1517249 


64 


3.1365485 


3.1441184 


128 


3.1403311 


3.1422236 


256 


3.1412772 


3.1417504 


512 


3.1415138 


3.1416321 


1024 


3.1415729 


3.1416025 


2048 


3.1415877 


3.1415951 


4096 


3.1415914 


3.1415933 . 


8192 


3.1415923 


3.1415928 


16384 


3.1415925 


3.1415927 



Now, the areas of the last two polygons differ fiom each 
other by less than the millionth part of the measuring unit. But 
the area of the circle differs from either, by less than they differ 
from each other ; hence, the value of the area of either will differ 
from that of the circle by less than a millionth part of the mea- 
suring unit. Taking the figures as far as they agree, and de- 
notino- the number of units in the required area by <tt, we have, 

JO 

approximately, 

* — 3.141592 ; 

that is, the area of a circle whose radius is 1, is 3.141592. 

Scholium. For practical computation, the value of tf is 
taken equal to 3.1416. 



BOOK V. 



153 



PROPOSITION XIII. THEOREM. 

The circumferences of circles are to each other as their radii, 
and the areas are to each other as the squares of their 
radii. 

Let C and be the centres of two circles whose 
radii are CA and OB : then will the circumferences be 
to each other as their radii, and the areas will be- to each 
other as the squares of their radii. 



M ^-^ N 




For, let similar regular polygons MNTJST and JEFGKL 
be inscribed in the circles : then will the perimeters of these 
polygons be to each other as their apothems, and the areas 
will be to each other as the squares of their apothems, what- 
ever may be the number of their sides (P. IX.). 

If the number of sides be made infinite (P. X. S. 2.), the 
polygons will coincide with the circles, the perimeters with 
the circumferences, and the apothems with the radii : hence, 
the circumferences of the circles arc to eacli other as their 
radii, and the areas are to each other as the squares of tho 
radii ; which was to be proved. 

Cor. 1. Diameters of circles are proportional to their 
radii : hence, the circumferences of circles are proportional 
to their diameters, and the areas are proportional to the 
squares of the diameters. 



154 



GEOMETRY. 




Cor. 2. Similar arcs, as AB and BE, are like parts 
of the circumferences to which 
they belong, and similar sectors, 
as A CB and JD OE, are like 
parts of the circles to which 
they belong : hence, similar 
arcs are to each other as their 
radii, and similar sectors are 
to each other as the squares of their radii. 

Scholium. The term infinite, used in the proposition, is to 
be understood in its technical sense. When it is proposed to 
make the number of sides of the polygons infinite, by the 
method indicated in the scholium of Proposition X., it is sim- 
ply meant to express the condition of things, when the in- 
scribed polygons reach their limits; in which case, the dif- 
ference between the area of either circle and its inscribed 
polygon, is less than any appreciable quantity. We have seen 
(P. XII.), that when the number of sides is 16384, the areas differ 
by less than the millionth part of the measuring unit. By increas- 
ing the number of sides, we approximate still nearer. 



PROPOSITION XIV. THEOREM. 

The area of a circle is equal to half the product of its 

circumference and radius. 

Let O be the centre of a circle, OC its radius, and 
A CDE its circumference : then will 
the area of the circle be equal to half 
the product of the circumference and 

radius. 

For, inscribe in it a regular poly- 
gon ACBK Then will the area of 
this polygon be equal to half the pro- 




BOOK V. 155 

duct of its perimeter and apothem, whatever may be the 
number of its sides (P. VIII.). 

If the number of sides be made infinite, the polygon will 
coincide with the circle, the perimeter with the circumference, 
and the apothem with the radius : hence, the area of the 
circle is equal to half the product of its circumference and 
radius ; which was to be proved. 

Cor. 1. The area of a sector is equal to half the pro- 
duct of its arc and radius. 

Cor. 2. The area of a sector is to the area of the circle, 
as the arc of the sector to the circumference. 

PROPOSITION XV. PROBLEM. 

To find an expression for the area of any circle in terms 

of its radius. 

Let C be the centre of a circle, and CA its radius. 
Denote its area by area CA, its radius 
by 12, and the area of a circle whose 
radius is 1, by < (P. XII., S.). 

Then, because the areas of circles 
are to each other as the squares of A 
their radii (P. XIII.), we have, 

. area CA : <t : : B 2 : 1 ; 
whence, area CA = tiW-. 

That is, the area of any circle is 3.1410 times the square 
of the radius. 

PROPOSITION XVI. PROBLEM. 

To find an expression for the circumference of a circle, in 
terms of its radius, or diameter. 

Lot C be the centre of a circle, and CA its radius. 





156 GEOMETRY. 

Denote its circumference by circ. GA, its radius by ii, atad 
its diameter by X>. From the last Proposition, we have, 

area GA = «B 2 ; 

and, from Proposition XIV., 'we have, 

area GA = \circ. G A x B ; A 

hence, \circ. GA x B = «B 2 ; 

whence, by reduction, 

circ. GA = 2tfB, or, circ. GA = <xD. 

That is, the circumference of any circle is equal to 3.1416 
times its diameter. 

Scholium 1. The abstract number tf, equal to 3.1416, de- 
notes the number of times that the diameter of a circle is 
contained in the circumference, and also the number of times 
that the square constructed on the radius is contained in the 
area of the circle (P. XV.). ISTow, it has been proved by 
the methods of Higher Mathematics, that the value of tt is 
incommensurable with 1 ; hence, it is impossible to express, 
by means of numbers, the exact length of a circumference 
in terms of the radius, or the exact area in terms of the 
square described on the radius. We may also infer that it 
is impossible to square the circle ; that is, to construct a 
square whose area shall be exactly equal to that of the cir 
cle. 

Scholium 2. Besides the approximate value of tf, 3.1416 
usually employed, the fractions ^ and fff are also used, 
when great accuracy is not required. 



BOOK VI. 

PLANES AND POLTEDEAL ANGLES. 

DEFINITIONS. 

1. A straight line is perpendicular to a plane, when 
it is perpendicular to every line of the plane which passes 
through its foot ; that is, through the point in which it 
meets the plane. 

In this case, the plane is also perpendicular to the line. 

2. A straight line is parallel to a plane, when it can- 
not meet the plane, how far soever both may be produced. 

In this case, the plane is also parallel to the line. 

3. Two Planes ape parallel, when they cannot meet, 
how far soever both may be produced. 

• 4. A Diedral angle is the amount of divergence of two 
planes. 

The line in which the planes meet, is called the edge of 
the angle, and the planes themselves are called faces of the 
angle. 

The measure of a diedral angle is the same as that of 
a plane angle formed by two lines, one drawn in each face, 
and both perpendicular to the edge at the same point. A 
diedral angle may be acute, obtuse, or a right angle. In 
the latter case, the faces are pcrpcridimlar to each other. 



158 



GEOMETRY. 



5. A Polyedkal angle is the amount of divergence of 
several planes meeting at a common point. 

This point is called the vertex of the angle ; the lines in 
which the planes meet are called edges of the angle, and 
the portions of the planes lying between the edges are 
called faces of the angle. Thus, 8 
i3 the vertex of the polyedral angle, 
whose edges are 8 A, 8JB, SC, 
SD, and whose faces are ASJB, 
BSC, CSD, DSA. 

A polyedral angle which has but 
three faces, is called a triedral 
angle. 




POSTULATE. 



A line may be drawn perpendicular to a plane from any 
point of the plane, or from any point without the plane. 



PROPOSITION I. THEOREM. 



If a straight line has two of its points in a plane, it tcitt 

lie wholly in that plane. 

For, by definition, a plane is a surface such, that if any 
two of its points be joined by a straight line, that line will 
lie wholly in the surface (B. I., D. 8). 

Cor. Through any point of a plane, an infinite number 
of straight lines may be drawn which will lie in the plane. 
-•For, if a line be drawn from the given point to any other 
point of the plane, that line will lie wholly in the plane. 

Scholium. If any two points of a plane be joined by a 
straight line, the plane may be turned about that line as an 




BOOK VI. 159 

axis, so as to take an infinite number of positions. Hence, 
we infer that an infinite number of planes may be passed 
through a given line. 

PROPOSITION H. THEOEEM. 

Through three points, not in the same straight line, one 
plane can be passed, and only one. 

Let A, B, and C be the three points : then can one 
plane be passed through them, and only one. 

Join two of the points, as A and 
B, by the line AB. Through AB 
let a plane be passed, and let this plane 
be turned around AB until it contains 
the point C ; in this position it will 
pass through the three points A, B, 
and C. If now, the plane be turned 

about AB, in either direction, it will no longer contain the 
point C : hence, one plane can always be passed through 
three points, and only one ; which icas to be proved. 

Cor. 1. Three points, not in a straight line, determine the 
position of a plane, because only one plane can be passed 
through them. 

Cor. 2. A straight line and a point without that line, 
determine the position of a plane, because only one plane 
can be passed through them. 

Cor. 3. Two straight lines which intersect, determine th 
position of a plane. For, let AB and AC intersect at 
A : then will either line, as AB, and one point of the 
other, as C, determine the position of a plane. 

Cor. 4. Two parallel lines determine the position of a 



160 



GEOMETRY. 



plane. For, let AB and CD be parallel. By definition 
(B. L, D. 16) two parallel lines always lie in the same plane. 
But either line, as AB, and any point 

of the other, as F, determine the posi- A _i> 

tion of a plane : hence, two parallels 

determine the position of a plane. E 



D 



PROPOSITION III. 



THEOEEM. 



The intersection of two planes is a straight line. 

Let AB and CD be two planes : then will their inter- 
section be a straight line. 

For, let E and F be any two 
points common to the planes ; draw 
the straight line EF. This line hav- 
ing two points in the plane AB, 
will lie wholly in that plane ; and 
having two points in the plane CD, 

will he wholly in that plane : hence, every point of EF is 
common to both planes. Furthermore, the planes can have 
no common point lying without EF, otherwise there would 
be two planes passing through a straight line and a point 
lying without it, which is impossible (P. II., C. 2) ; hence, 
the intersection of the two planes is a straight line ; which 
was to be proved. 




PROPOSITION IY. THEOEEM. 

If a straight line is perpendicular to two straight lines at 
their point of intersection, it is perpendicular to the plane 
of those lines. 

Let MN~ be the plane of the two lines BB, CC, and 
let AB be perpendicular to these lines at P : then will 



BOOK VI. 



161 



AP be perpendicular to every line of the plane which 
passes through P, and consequently, to the plane itself. 

For, through P, draw in 
the plane MN, any line PQ ; 
through any point of this line, 
as Q, draw the line BC, so 
that BQ shall be equal to QC 
(B. IV., Prob. V.) ; draw AB, 
AQ, and AC. 

The base BC, of the triangle BPC, being bisected at 
Q, we have (B. IV., P. XIV.), 




PC + PB* = 2PQ* + 2QC\ 

In like manner, we have, from the triangle ABC, 

AC 2 + AB 2 = 2AQ 2 + 2 QC 2 . 

Subtracting the first of these equations from the second, 
member from member, we have, 



AC - PC + AIT - PB' = 2AQ' - 2PQ\ 

But, from Proposition XI., C. 1, Book IV., we have, 

AC 2 - PC 2 = AP 2 , and AB 2 - PB 2 = AP 2 ; 
hence, by substitution, 



whence, 



2AP' = 2AQ* - 2PQ', 



or, 



AP 2 + P© 2 = A Q\ 



AP 2 = AQ 2 - PQ 2 ; 

The triangle APQ is, therefore, right-angled at P (B. IV., 
P. XIII., S.), and consequently, .4JP is perpendicular to 
PQ : hence, AP is perpendicular to every line of the 
plane MJ¥ passing through P, and consequently, to the 
plane itself ; which was to be proved. 

11 



162 



GEOMETRY. 




Cor. 1. Only one perpendicular can be drawn to a plane 
from a point without the plane. 
For, suppose two perpendiculars, 
as AP and AQ, could be 
drawn from the point A to the 
plane 31JSF. Draw PQ ; then 
the triangle APQ would have 
two right angles, APQ and 
A QP ; which is impossible (B. I., P. XXV., C. 3). 

Cor. 2. Only one perpendicular can be drawn to a plane 
from a point of that plane. For, suppose that two perpen- 
diculars could be drawn to the plane JJilV, from the point 
P. Pass a plane through the perpendiculars, and let PQ 
be its intersection with 3IJSF; then we should have two per- 
pendiculars drawn to . the same straight line from a point of 
that line ; which is impossible (B. I., P. XIV., C). 



1° 

2° 



PROPOSITION V. THEOREM. 

If from a point loithout a plane, a perpendicular he drawn 
to the plane, and oblique lines be drawn to different 
points of the plane : 
The perpendicular icill be shorter than any oblique line : 
Oblique lines tohich meet the plane at equal distances 
from the foot of the perpendicular, will be equal: 
3.° Of tioo oblique lines which meet the plane at unequal 
distances from the foot of the perpendicular, the one which 
meets it at the greater distance will be the longer. 

Let A be a point without the plane 3IN ; let Al 
be perpendicular to the plane ; let A. C, AP, be any two 
oblique lines meeting the plane at equal distances from the 
foot of the perpendicular ; and let A C and AE be any 




BOOK VI. 163 

two oblique lines meeting the plane at unequal distances from 
the foot of the perpendicular : 

1°. AP will be shorter 
than any oblique line AG. 

For, draw PC; then will 
AP be less than AG (B. 
I., P. XV.) ; which was to 
be proved. 

2°. AG and AD will be equal. 

For, draw PD ; then the right-angled triangles APC, 

APD, will have the side AP common, and the sides PC, 

PD, equal : hence, the triangles are equal in all their parts, 

and consequently, A G and AD will be equal ; which was 

to be proved. 

• 

3°. AE will be greater than AG. 

For, draw PE, and take PD equal to PG ; draw 
AD : then will AE be greater than AD (B. I., P. XV.) ; 
but AD and A G are equal : hence, AE is greater than 
AG ; lohich was to be proved. 

Cor. The equal oblique lines AD, AG, AD, meet the 
plane MIT in the circumference of a circle, whose centre ia 
P, and whose radius is PD : hence, to draw a perpendi- 
cular to a given plane MIT, from a point A, without that 
plane, find three points D, G, D, of the plane equally dis- 
tant from A, and then find the centre P, of the circle 
whose circumference passes through these points : then will 
AP be the perpendicular required. 

Scholium. The angle ADP is called the inclination of 
the oblique line AD to the plane MIT. The equal oblique 
lines AD, AG, AD, are all equally inclined to the plane 
MIT. The inclination of AE is less than the inclination of 
any shorter line AD. 



164 



GEOMETRY. 



PROPOSITION YI. THEOREM. 

If from the foot of a 'perpendicular to a plane, a line be 
drawn at right angles to any line of that plane, and the 
point of intersection be joined with any point of the per- 
pendicular, the last line will be perpendicular to the line 
of the plane. 

Let AP be perpendicular to the plane MN, P its foot, 
B C the given line, and A any point of the perpendicular ; 
draw PP at right angles to PC, and join the point P 
with A : then will AP be perpendicular to PC. 

For. lay off PP equal to 
PC, and draw PP, PC, AP, 
and AC. Because PP is per- 
pendicular to PC, and PP 
equal to PC, we have, PP 
equal to PC (B. I, P. XV.) ; 
and because AP is perpendicu- 
lar to the plane MN", and PP 

equal to PC, we have AP equal to AC (P. V.). The 
line AP has, therefore, two of its points A and P, each 
equally distant from P and C : hence, it is perpendicular 
to PC (B. I., P. XVI., S.) ; which was to be proved. 

Cor. 1: The line PC is perpendicular to the plane of 
the triangle A PP ; because it is perpendicular to AP and 

PP, at P (P. rv.). 

Cor. 2. The shortest distance between AP and PC is 
measured on PP, perpendicular to both. For, draw PP 
between any other points of the lines : then will PP be 
greater than PP, and PP will be greater than PP : 
hence, PP is less than PP. 




BOOK VI. 



165 



Scholium. The lines AP and DC\ though not in the 
same plane, are considered perpendicular to each other. In 
genera], any two straight lines not in the same plane, are 
considered as making an angle with each other, which angle 
is equal to that formed by drawing through a given point, 
two lines respectively parallel to the given lines. 




PROPOSITION VII. THEOREM. 

If one of two parallels is perpendicular to a plane, the other 
one is also perpendicular to the same plane. 

Let AP and ED be two parallels, and let AP be 
perpendicular to the plane MJST : then will ED be also 
perpendicular to the plane MN~. 

For, pass a plane through the 
parallels ; its intersection with 
MJST will be PD ; draw A.D, 
and in the plane 3IJV draw 
DC perpendicular to PD at 
D. Now, DD is perpendicular 
to the plane APDE (P. VI., C.) ; 

the angle DDE is consequently a right angle ; but the an- 
gle EDP is a right angle, because ED is parallel to AP 
(B. I., P. XX., C. 1) : hence, ED is perpendicular to DD 
and PD, at their point of intersection, and consequently, to 
'their plane 3IJST (P. IV.) ; which was to be proved. 

Cor. 1. If the lines AP and ED are perpendicular to 
the plane MN~, they are parallel to each other. For, if 
not, draw through D a line parallel to PA ; it will bo 
perpendicular to the plane MN, from what has just been 
proved ; we shall, therefore, have two perpendiculars to the 
the plane MN~, at the same point ; which is impossible (P. 
IV., C. 2). 



166 



GEOMETRY. 



Cor. 2. If two lines, A and B, are parallel to a third 
line (7, they are parallel to each other. For, pass a plane 
perpendicular to C ; it will be perpendicular to both A 
and B : hence, A and B are parallel. 




B 



PROPOSITION VIII. THEOREM. 

If a line is parallel to a line of a plane, it is parallel to 

that plane. 

Let the line AB be parallel to the line' CD of the 
plane MJV ; then will AB be parallel to the plane MN". 

For, through AB and CD 
pass a plane (P. II., C. 4) ; CD 
will be its intersection with 
the plane MN. ISTow, since AB 
lies in this plane, if it can meet 
the plane MN, it will be at 
some point of CD ; but this is 

impossible, because AB and CD are parallel : hence, AB 
cannot meet the plane MN, and consequently, it is parallel 
to it ; which was to he proved. 




PROPOSITION IX. THEOREM. 

If two planes are perpendicular to the same straight line. 

they are parallel to each other. 

Let the planes MN" and PQ 
be perpendicular to the line AB, 

at the points A and B : then O 

will they be parallel to each 
other. 

For, if they are not parallel, 







Q 


p/ 


B 


r--/. 


1 

M 


- 


/ A ' 


"'7 






$ 



BOOK VI. 



167 



they will meet ; and let be a point common to both. 
From draw the lines OA and OB : then, since OA 
lies in the plane JIJV, it will be perpendicular to BA at 
A (D. 1). For a like reason, OB will be perpendicular 
to AB at B : hence, the triangle OAB will have two 
right angles, which is impossible ; consequently, the planes 
cannot meet, and are therefore parallel ; which was to be 
proved. 

PROPOSITION X. THEOREM. 



If a plane, intersect tico parallel planes, the lines of inter- 
section will be parallel. 

Let the plane EII intersect the parallel planes 3IJSF and 
PQ, in the lines EF and GH : then will EF and GH 
be parallel. 

For, if they are not parallel, 
they will meet if sufficiently })ro- 
longed, because they lie in the 
same plane ; but if the lines meet, 
the planes 3IJST and P Q, in 
which they lie, will also meet ; 
but this is impossible, because 
these planes are parallel : hence, 
the lines EF and Gil cannot meet 
parallel ; which was to be proved. 




they are, therefore, 



PROPOSITION XI. THEOREM. 

If a straight line is perpendicular to one of two parallel 
planes, it is also perpendicular to the other. 

Let MN and PQ be two parallel planes, and let the 
line AD be perpendicular to PQ then will it also be 

perpendicular to 3IJST. 



168 



GEOMETRY. 



Q 



For, through AB pass any plane ; its intersections with 
3IJST and PQ will be parallel (P. X.) ; but, its intersection 
with PQ is perpendicular to AB at B (D. 1) ; hencfc. 
its intersection with 3IN~ is 
also perpendicular to AB at A 
(B. I, P. XX., C. 1) : hence, 
AB is perpendicular to every 
line of the plane 3IN" through 
A, and is, therefore, perpendicu- 
lar to that plane ; which was to 
be proved. 



/ B 


\ / 


p 


N 


/ * 


</ 



M 



PROPOSITION XII. THEOREM. 
Parallel lines included between parallel planes, are equal. 

Let EG and FPI be any two parallel lines included 
between the parallel planes 3IJST and PQ : then will they 
be equal. 

Through the parallels conceive 
a 'plane to be passed ; it ' will 
intersect the plane MN in the 
line EI] and PQ in the line 
Gil ; and these lines will be 
parallel (Prop. X.). The figure 
EFHG is, therefore, a parallelo- 
gram : hence, GE and IIF 
are equal (B. I., P. XXVIII.) ; lohich was to be proved. 

Cor. 1. The distance between two parallel planes is mea- 
sured on a perpendicular to both ; but any two perpendiculars 
between the planes are equal : hence, parallel planes are every- 
where equally distant. 

Cor. 2. If a line Gil is parallel to any plane MN, 

then can a plane be passed through Gil parallel to 3£N~ : 

hence, if a line is parallel to a plane, all of its points are 
equally distant from that plane. 




BOOK VI. 



169 



proposition xni. theorem 

If two angles, not situated in the same plane, have their 
sides parallel and lying in the same direction, t/ie angles 
will be equal and their planes parallel. 

Let CAE''tm<\. DBF be two angles lying in the planes 
MJST and PQ, and let the sides AC and AE be re- 
spectively parallel to BD and BE, and lying hi the same 
direction : then will the angles CAE and DBF be equal, 
and the planes MN and PQ will be parallel. 

Take any two points of AC and AE, as C and E, and 
make BD equal to AC, and 
BE to AE ; draw CE, DF, 
AB, CD, and EF 

1°. The angles CAE and 
DBF will be equal. 

For, AE and BF being 
parallel and equal, the figure 
ABFE is a parallelogram (B. 
L, P. XXX.) ; hence, EF is 
parallel and equal to AB. For 
a like reason, CD is parallel and equal to AB : hence, 
CD and EF are parallel and equal to each other, and 
consequently, CE and DF are also parallel and equal to 
each other. The triangles CAE and DBF have, therefore, 
their corresponding sides equal, and consequently, the corres- 
ponding angles CAE and DBF are equal ; which was to 
be proved. 

2°. The planes of the angles MN and PQ are parallel. 

For, if not, pass a plane through A parallel to PQ, 
and suppose it to cut the lines CD and EF in G and 
H. Then will the lines CD and 1IF be equal respect- 




170 



GEOMETRY. 



ively to AB (P. XII.), and consequently, GD will be 

equal to CD, and JECF to EF\ which is impossible : hence, 

the planes MN and JPQ must be parallel ; which was to 

be proved. 

Cor. If two parallel planes MN and PQ, are met by 
two other planes AD and AF, the angles CAF and 
DBF, formed by their intersections, will be equal. 



PROPOSITION XIV. THEOREM. 



If three straight lines, not situated in the same plane, are 
equal and parallel, the triangles formed by joining the 
extremities of these lines will be equal, and their planes 

parallel. 

Let AB, CD, and EF be equal parallel lines not in 
the same plane : then will the triangles A CF and BDF 
be equal, and their planes parallel. 

For, AB being equal and 
parallel to EF, the figure ABFE 
is a parallelogram, and conse- 
quently, AE is equal and par- 
allel to BE. For a like reason, 
AC is equal and parallel to 
BD : hence, the included angles 
CAE and DBF are equal and 
their planes parallel (P. XIII.). 
Now, the triangles CAE and 
DBF have two sides and their 

included angles equal, each to each : hence, they are equal 
in all their parts. The triangles are, therefore, equal and 
their planes parallel ; which was to be proved. 




BOOK VI. 



171 



PROPOSITION XY. THEOEEM. 

If two straight lines are cut by three parallel planes, they 

will be divided proportionally. 

Let the lines AB and CB be cut by the parallel 
planes JUST, PQ, and US, in the points A, E, B, and 
G, F, B; then 



AE 



EB 



CF : FB. 




For, draw the line AB, and 
suppose it to pierce the plane 
P Q in G ; draw A C, BB, 
EG, and GF. 

The plane ABB intersects 
the parallel planes BS and PQ 
in the lines BB and EG ; 
consequently, these lines are par- 
allel (P. X.) : hence (B. IV., 
P. XV.), 

AE : EB : : AG : GB. 

The plane A CB intersects the parallel planes MN and 
PQ, in the parallel lines AG and GF : hence, 

AG : GB : : CF : -ED. 

Combining these proportions (B. II., P. IV.), we have, 

AE : EB : : CF : FB ; 

which was to be proved. 

Cor. 1. If two lines are cut by any number of parallel 
planes they will be divided proportionally. 

Cor. 2. If any number of lines are cut by three parallel 
planes, they will be divided proportionally. 



172 GEOMETRY. 

PROPOSITION XVI. THEOREM. 

Tf a line is perpendicular to a plane, every plane passed 
through the line will also he perpendicular to that plane, 

Let AP be perpendicular to the plane MN, and let 
BF be a plane passed through AP : then will BF be 
perpendicular to 31N". 

In the plane 3Tj¥, draw PD 
perpendicular to PC, the intersec- 
tion of BF and 3IJV. Since AP r 
is perpendicular to MIST, it is per- / 
pendicular to BG and DP (D. 1); / » - -p/ 
and since AP and BP, in the 

planes BF and 3IJV, are perpendicular to the intersection 
of these planes at the same point, the angle which they 
form is equal to the angle formed by the planes (D. 4) ; 
but this angle is a right angle : hence, BF is perpendicu- 
lar to 3IJV ; ichich was to be proved. 

Cor. If three lines AP, BP, and BP, are perpen- 
dicular to each other at a common point P, each line will 
be perpendicular to the plane of the other two, and the 
three planes will be perpendicular to each other. 

PROPOSITION XVII. THEOREM. 

If two planes are perpendicular to each other, a line drawn 
in one of them, p>erpendicidar to their intersection, loiU 
be perpendicular to the other. 

Let the planes BF and MJV be perpendicular to each 
other, and let the line AP, drawn in the plane BF, be 
perpendicular to the intersection BC ; then will AP be 
perpendicular to the plane 3IN. 



BOOK VI. 



173 




For, in the plane MN, draw PD perpendicular to PC 

at P. Then because the planes PF and MN are perpen- 
dicular to each other, the angle APP 
will he a right angle : hence, AP is 
perpendicular to the two lines PD 
and PC, at their intersection, and 
consequently, is perpendicular to their 
plane MN ; which icas to be proved. 

Cor. If the plane PF is perpendicular to the plane 
MN, and if at a point P of their intersection, Ave erect 
a perpendicular to the plane MN, that perpendicular will 
be in -the plane PF. For, if not, draw in the plane PF, 
PA perpendicular to PC, the common intersection ; AP 
will be perpendicular to the plane MJST, by the theorem ; 
therefore, at the same point P, there are two perpendiculars 
to the plane MN ; which is impossible (P. IV., C. 2). 



PROPOSITION XVIII. THEOREM. 



If ttoo planes cut each other, and are perpendicular to a 
third plane, their intersection is also 2nrpendicxdar to 
that plane. 

Let the planes PF, PIT, be perpendicular to MN : 
then will their intersection AP be perpendicular to MN. 

For, at the point P, erect a per- 
pendicular to the plane MN ; that 
perpendicular must be in the plane 
PF, and also in the plane PIT 
(P. XVII., C.) ; therefore, it is their 
common intersection AP : which icas 
to be proved. 



M 




TV^i'^Jn 



N 



174: GEOMETRY. 

PKOPOSITION XIX. THEOEEM. 

The sum of any two of the plane angles formed by the 
edges of a triedral angle, is greater than the third. 

Let SA, SB, and SG, be the edges of a triedral 
angle : then will the sum of any two of the plane angles 
formed by them, as ASG and GSB, be greater than the 

third A SB. 

If the plane angle A SB is equal to, or less than, either 
of the other two, the truth of the proposition is evident. Let 
us suppose, then, that A SB is greater than either. 

In the plane A SB, construct 
the angle BSD equal to BSG ; 
draw AB in that plane, at plea- 
sure ; lay off SO equal to SB, 
and draw AG and CB. The 
triangles BSB and BSG have 
the side SG equal to SB, by 
construction, the side SB com- 
mon, and the included angles BSB and BSG equal, by 
construction ; the triangles are therefore equal in all their 
parts : hence, BB is equal to BG. But, from Proposition 
VII., Book I., we have, 

BG + GA>BB + BA. 

Taking away the equal parts BG and BB, we have, 

GA > BA ; 

hence (B. I., P. IX., C), we have, 

angle ASG > angle A SB ; 

and, adding the equal angles BSG and BSB, 




BOOK VI. 



175 



or, 



angle ASC -f angle CSB > angle A SB + angle BSB ; 
angle ASC + angle CSB > angle A SB ; 



which was to be proved. 



PROPOSITION XX. THEOREM. 

The sum of the plane angles formed by the edges of any 
polyedral angle, is less than four right angles. 

Let S be the vertex of any polyedral angle whose edges 
are SA, SB, SC, SB, and SE ; then will the sum of 
the angles about S be less than four right angles. 

For, pass a plane cutting the edges 
in the points A, B, C, B, and E, 
and the faces in the lines AB, B C, 
CB, BE, and EA. From any point 
within the polygon thus formed, as 0, 
draw the straight lines OA, OB, OC, 
OB, and OE. 

We then have two sets of triangles, 
one set having a common vertex S, the 
other having a common vertex 0, and both having com- 
mon bases AB, BC, CB, BE, EA. Now, in the set 
which has the common vertex S, the sum of all the angles 
is equal to the sum of all the plane angles formed by the 
edges of the polyedral angle whose vertex is S, together 
with the sum of all the angles at the bases : viz., SAB, 
SB A, SBC, &c. ; and the entire sum is equal to twice 
as many right angles as there are triangles. In the set 
whose common vertex is 0, the sum of all the angles is 
equal to the four right angles about 0, together with the 
interior angles of the polygon, and this sum is equal to 
twice as many right angles as there are triangles. Since 




176 



GEOMETRY. 



the number of triangles, in each set, is the same, it follows 
that these sums are equal. But in the triedral angle whose 
vertex is B, we have (P. XIX.), 

ABS + SBC > ABC y 

and the like may be shown at each 

of the other vertices, C, B, E, A : 

hence, the sum of the angles at the 

bases, in the triangles whose common 

vertex is S, is greater than the sum 

of the angles at the bases, in the set 

whose common vertex is : therefore, 

the sum of the vertical angles about S, is less than the 

sum of the angles about : that is, less than four right 

angles ; which loas to be proved. 

Scholium. The above demonstration is made on the sup- 
position that the polyedral angle is convex, that is, that the 
diedral angles of the consecutive faces are each less than two 
right angles. 




PKOPOSITION XXI. 



THEOKEM. 



If the plane angles formed by the edges of two triedral 
angles are equal, each to each, the planes of the equal 
angles are equally inclined to each other. 

Let S and T be the vertices of two triedral angles, 
and let the angle ASC be equal to DTF, ASB to BTE, 
and BSC to ETF : then will the planes of the equal 
angles be equally inclined to each other. 

For, take any point of SB, as B, and from it draw 
in the two laces ASB and CSB, the lines BA and BC, 
respectively perpendicular to SB : then will the angle ABC 
measure the inclination of these faces. Lay off TE .equal' 



BOOK VI. 



177 




to SB, and from E draw in the faces DTE and FTE, 
the lines ED and EF, respectively perpendicular to TE • 
then will the angle DEE 
measure the inclination of these 
faces. Draw AC and DF. 
The right-angled triangles 
SBA and TED, have the 
side SB equal to TE, and 
the aDgle A SB equal to 
DTE; hence, AB is equal to DE, and .4/S to TD. 
In like manner, it may he shown that BC is equal to .EF, 
and CS to .FT! The triangles ASO and X>77^ have 
the angle ASC equal to Z>ri^ by hypothesis, the side AS 
equal to DT, and the side CS to i^Z 7 , from what has 
just been shown ; hence, the triangles are equal in all their 
parts, and consequently, AC is equal to DF. Xow, the 
triangles ABC and DEE have their sides equal, each to 
each, and consequently, the corresponding angles are also 
equal; that is, the angle ABC is equal to DEE: hence T 
the inclination of the planes ASB and CSB, is equal to 
the inclination of the planes DTE and FTE. In like 
manner, it may be shown that the planes of the other equal 
angles are equally inclined ; which was' to be proved. 

Scholium. If the planes of the equal plane angles are 
like placed, the triedral angles are equal in all respects, for 
they may be placed so as to coincide. If the planes of the 
equal angles are not similarly placed, the triedral angles are 
equal by symmetry. In this case, they may be placed so 
that two of the homologous faces shall coincide, the triedral 
angles lying on opposite sides of the plane, which is then 
called a plane of symmetry. In this position, for every point 
on one side of the plane of symmetry, there is a correspond- 
ing point on the other side. 

12 



OK VII. 



POLYEDEONS. 



DEFINITIONS. 



1. A Polyedeon is a volume bounded by polygons. 

The bounding polygons are called faces of the polyedron; 
the lines in which the polygons meet, are called edges of the 
polyedron; the points in which the edges meet, are called 
vertices of the polyedron. 

2. A Peism is a polyedron, two of whose 
faces, are equal polygons having their homo- 
logous sides parallel, the other faces being 
parallelograms. 

The equal polygons are called bases of the 
prism ; one the upper, and the other the 
lower base ; the parallelograms taken together 
make up the lateral or convex surface of the prism 
lines in which the lateral faces meet, are called lateral 
of the prism. 




; the 

edges 



3. The Altitude of a prism is the perpendicular dis- 
tance between the planes of its bases, 

4. A Right Peism is one whose lateral 
edges are perpendicular to the planes of the 

bases, 

In this case, any lateral edge is equal to 

the altitude. 




BOOK VII. 



179 



5. An Oblique Prism is one whose lateral edges are 
oblique to the planes of the bases. 

In this case, any lateral edge is gi eater than the altitude. 

6. Prisms are named from the number of sides of their 
bases ; a triangular prism is one whose bases are triangles ; 
a quadrangular prism is one whose bases are quadrilaterals ; 
a pentangular prism is one whose bases are pentagons, and 
so on. 

7. A Parallelopipedon is a prism whose bases are 
parallelograms. 

A Rectangular Parallelopipedon is a right 
parallelopipedon, all of whose faces are rect- 
angles ; a cube is a rectangular parallelo- 
pipedon, all of whose faces are squares. 



8. A Pyramid is a polyedron bounded 
by a polygon called the base, and by tri- 
angles meeting at a common point, called the 
vertex of the pyramid. 

The triangles taken together make up the 
lateral or convex surface of the pyramid ; 
the lines in which the lateral faces meet, are 
called the lateral edges of the pyramid. 



9. Pyramids are named from the number of sides of 
their bases ; a triangular pyramid is one whose base is a 
triangle ; a quadrangular pyramid is one whoso base is a 
quadrilateral, and so on. 

10. The Altitude of a pyramid is the perpendicular 
distance from the vertex of the pyramid to the plane of its 
base. 



< . 



\ 

\ 




\ 


1 

1 

....... ^ 

\ 





180 GEOMETRY. 

11. A Right Pyramid is one whose base is a regular 
polygon, and in which the perpendicular drawn from the 
vertex to the plane of the base, passes through the centre 
of the base. 

This perpendicular is called the axis of the pyramid. 

12 The Slant Height of a right pyramid, is the per- 
pendicular distance from the vertex to any side of the base. 

13. A Truncated Pyramid is that 
portion of a pyramid included between 
the base and any plane which cuts the 
pyramid. 

When the cutting plane is parallel to 
the base, the truncated pyramid is called 
a frustum op a pyeamid, and the inter- 
section of the cutting plane with the pyramid, is called the 
upper base of the frustum ; the base of the pyramid is cal- 
led the loioer base of the frustum. 

14. The Altitude of a frustum of a pyramid, is the per- 
pendicular distance between the planes of its bases. 

15. The Slant Height of a frustum of a right pyramid, 
is that portion of the slant height of the pyramid which lies 
between the planes of its upper and lower bases. 

16. Similar Polyedeons are those which are bounded by 
similar polygons, similarly placed. 

Parts which are similarly placed, whether faces, edges, or 
angles, are called homologous. 

17. A Diagonal of a polyedron, is a straight line join- 
ing the vertices of two polyedral angles not in the same 

face. 



BOOK VII. 



181 



18. The Volume of a Polyeduon is its numerical value 
expressed in terms of some other polyedron as a unit. 

The unit generally employed is a cube constructed on tlia 
linear unit as an ed^e. ' 



PKOPOSITIOiN" I. TBEOEEM. 



The convex surface of a right prism is equal to the perim- 
eter of either base multiplied by the altitude. 

Let ABGDE-K be a right prism : then is its convex 
surface ' equal to, 

(AB + JBC + CD + BE + EA) x AF. 



For, the convex surface is equal to 
the sum of all the rectangles AG, BIT, 
CI, BE, EF, which compose it. Now, 
the altitude of each of the rectangles 
AF, BG, Gil, Sec, is equal to the 
altitude of the prism, and the area of 
each rectangle is equal to its base mul- 
tiplied by its. altitude (B. IV., P. V.) : 
hence, the sum of these rectangles, or 
the convex surface of the prism, is equal to, 



(AB + BG + CD 4- BE + EA) x AF ; 

that is, to the perimeter of the base multiplied by the alti- 
tude ; which was to be proved. 

Cor. If two right prisms have the same altitude, their 
convex surfaces are to each other as the perimeters of their 
bases. 




182 



GEOMETRY. 



PROPOSITION II. THEOREM. 

In any prism, the sections made by parallel planes are equal 

polygons. * 

Let the prism AH be intersected by the parallel planes 
NT, SV : then are the sections NOPQB, STVXY, 
equal polygons. 

For, the sides NO, ST, are parallel, 
being the intersections of parallel planes 
with a third plane ABGT; these sides, 
NO, ST, are included between the par- 
allels NS, OT: hence, NO is equal to 
ST (B. I., P. XXVni, C. 2). For like 
reasons, the sides OT, T Q, QB, &c, 
of NOTQB, are equal to the sides 
TV, VX, &c, of STVXY, each to 
each ; and since the equal sides are par- 
allel, each to each, it follows that the 

angles NOP, OT Q, &c, of the first section, are equal to 
the angles STV, TVX, &c, of the second section, each to 
each (B. VI., P. XIII.) : hence, the two sections NOTQB, 
STVXY, are equal polygons ; which icas to be proved. 

Cor. Every sectior of a prism, parallel to the bases, is 
equal to either base. 




PROPOSITION III. 



THEOREM. 



If a pyramid be cut by a plane parallel to the base ' 

1°. The edges and the altitude will be divided proportionally : 
2°, The section will be a polygon similar to the base. 

Let the pyramid S-ABCJDE, whose altitude is SO, 
be cut by the plane abcde, parallel to the base ABODE. 



BOOK VII. 



183 




lo, The edges and altitude will be divided proportionally. 

F02, conceive a plane to be passed through the vertex S, 
parallel to the plane of the base ; then 
will the edges and the altitude be cut 
by three parallel planes, and consequently 
they will be divided proportionally (B. VI., 
P. XV., C. 2) ; which was to be proved. 

2°. The section abode, will be similar 
to. the base ABCBE. For, ab is par- 
allel to AB, and be to BC (B. VI., 
P. X.) : hence, the angle abc is equal to 
the angle ABC. In like manner, it may 
be shown that each angle of the polygon abode is equal 
to the corresponding angle of the base : hence, the two 
polygons are mutually equiangular. 

Ao-ain, because ab is parallel to AB, we have, 

ab : AB : : sb : SB ; 
and, because be is parallel to BC, we have, 

be : BC : : sb : SB ; 
hence (B. II., P. IV.), we have, 

ab : AB : : be : BC. 

In like manner, it may be shown that all the sides of 

abode are proportional to the corresponding sides of the 

polygon ABCBE : hence, the section abode is similar to 

the base ABCBE (B. IV., D. 1) ; which was to be proved. 

Cor. 1. If two pyramids S- ABCBE, and S-XY? y 
having a common vertex S, and their bases in the same 
plane, be cut by a plane abc, parallel to the plane of 
their bases, the sections will be to each other as the bases. 



184: 



GEOMETRY. 



For, the polygons abed and ABCD, being similar, are 
to each other as the squares of their homologous sides ab 
and AB (B. IV., P. XXVII) ; but, 



So 



SO'; 



ab' : AB' : : Sa : SA l 



bance (B. II., P. IV.), we have, 



abode : ABODE : : So 2 : SO 2 . 



In like manner, we have, 

xyz : XYZ : : So : ^0 2 ; A ' 

hence, 

abode : ABODE : : xyz : XYZ. 

Cor. 2. If the bases are equal, any sections at equal dis- 
tances from the bases will be equal. 

Cor. 3. The area of any section parallel to the base, is 
proportional to the square of its distance from the vertex, 




PROPOSITION IV. 



THEOREM. 



The convex surface of a right pyramid is equal to tJie 
perimeter of its base multiplied by half the slant height. 

Let S be the vertex, ABCDE the 

base, and SE, perpendicular to EA, the 

slant height of a right pyramid : then will 
the convex surface be equal to, 

(AB + BC + CD + DE+ EA) x l&F. 

Draw SO perpendicular to the plane of the 
base. 




/ 



BOOK VII. 



185 



From the definition of a right pyramid, the point O is 
the centre of the base (D. 11) : hence, the lateral edges, 
SA, SB, &c, are all equal (B. VI., P. V.) ; but the sides 
of the base are all equal, being sides of a regular polygon : 
hence, the lateral faces are all equal, and consequently their 
altitudes are all equal, each being equal to the slant height 
of the pyramid. 

Now, the area of any lateral face, as SEA, is equal to 
its base FA, multiplied by half its altitude SF : hence, 
the sum of the areas of the lateral faces, or the convex sur- 
face of the pyramid, is equal to, 

. {AB + BO + CD + BE + EA) x \8F ; 

which was to be proved. 



Scholium. The convex surface of a frustum of a right 
pyramid is equal to half the sum of the perimeters of its 
upper and lower bases, multiplied by the slant height. 

Let ABCBE-e be a frustum of a right 
pyramid, whose vertex is S : then will the 
section abcde be similar to the base AB CDE, 
and their homologous sides will be parallel, 
(P. III.). Any lateral face of the frustum, 
as AEea y is a trapezoid, whose altitude is 
equal to Ff the slant height of the frustum ; 
hence, its area is equal to \ [EA + ea) x Ff 
(B. IV., P. VII.). But the area of the con- 
vex surface of the frustum is equal to the sum of the areas 
of its lateral faces ; it is, therefore, equal to the half sum 
of the perimeters of its upper and lower bases, multiplied 
by half the slant height. 




\ 



186 



GEOMETRY. 



PEOPOSITION V. THEOEEM. 

If the three, faces which include a triedral angle of a prism 
are equal to the three faces which include a triedral an- 
gle of a second prism, each to each, and are like placed, 
the two prisms are equal in all their parts. 

Let B and b be the vertices of two triedral angles, 
included by faces respectively equal to each other, and simi- 
larly placed: then will the prism ABCBE-K be equal to 
the prism abcde-Jc, in all of its parts. 

For, place the base 
alcde upon the equal 
base ABC BE, so that 
they shall coincide ; then 
because the triedral an- 
gles whose vertices are 
b and B, are equal, 
the parallelogram bh will 
coincide with BH, and 
the parallelogram bf with 
BF : hence, the two 

sides fg and gh, of one upper base, will coincide with the 
homologous sides of the other upper base ; and because the 
upper bases are equal, they must coincide throughout ; con- 
sequently, each of the lateral faces of one prism will coincide 
with the corresponding lateral face of the other prism: the 
prisms, therefore, coincide throughout, and are therefore equal 
in all their parts ; which was to be proved. 

Cor. If two right prisms have their bases equal in all their 
parts, and have also equal altitudes, the prisms themselves will 
be equal in all their parts. For, the faces which include any 
triedral angle of the one, will be equal to the faces which 
include the corresponding triedral angle of the other each to 
each, and they will be similarly i:>laced. 




BOOK VII. 



137 




PROPOSITION VI. THEOREM. 

In any parallelopipedon, the opposite faces are equal, each 
to each, and their planes are parallel. 

Let ABCD-H be a parallelopipedon : then will its 
opposite faces be equal and their planes will be parallel. 

For, the bases, ABCD and EFGH 
are equal, and their planes parallel by 
definition (D. 1). The opposite faces 
AEIID and BFGC, have the sides AE 
and BF parallel, because they are oppo- 
site sides of the parallelogram BE ; 
and the sides EII and EG parallel, 

because they are opposite sides of the parallelogram EG; 
and consequently, the angles AEII and BEG are equal 
(B. VI., P. XIII.). But the side AE is equal to BF, and 
the side EII to EG ; hence, the faces AEHD and 
BFGC are equal ; and because AE is parallel to BF, 
and EII to EG, the planes of the faces are parallel 
(B. VI., P. XIIL). In like manner, it may be shown that 
the parallelograms ABFE and DCGH, are equal and their 
planes parallel : hence, the opposite faces are equal, each to 
each, and their planes are parallel ; which was to be proved. 

Cor. 1. Any two opposite faces of a paraJlelopipedon 
may be taken as bases. 

Cor. 2. In a rectangular parallelo- 
pipedon, the square of either of the 
diagonals is equal to the sum of the 
squares of the three edges which meet 
at the same vertex. 

For, let EI) be either of the diagonals, and draw EII. 




188 



GEOMETRY. 



Then, in the right-angled triangle FHD, we have, 



FIT = DH* + Fir. 



But DII is equal to FB, and i^T 2 
is equal to FA 2 plus JZ£f 2 or FC 2 : 

hence, 




.EZr = FB A + .KT + -^ 



Cor. 3. A parallel opipedon may be constructed on three 
lines AB, AD, and AE, intersecting in a common point 
A, and not lying in the same plane. For, pass through 
the extremity of each line, a plane parallel to the plane of 
the other two lines ; then will these planes, together with 
the planes of the given lines, determine a parallelopipedon. 



PROPOSITION VII. THEOREM. 

If a plane be passed through the diagonally op>posite edges 
of a parallelopipedon, it will divide the parallelopipedon 
into two equal triangular prisms. 

Let ABCD-II be a parallelopipedon, and let a plane 
be passed through the edges BF and DII : then wihV»the 
prisms ABD-II and BCD-II be equal 
in volume. 

For, through the vertices F and B 
let planes be passed perpendicular to 
FB, the former cutting the other lateral 
edges in the points e, h, g, and the 
latter cutting those edges produced, in 
the points a, d, and c. The sections 
Fehg and Bade will be parallelograms, 




BOOK VII. 189 

because their opposite sides are parallel, each to each (B. VI., 
P. X.) ; they will also be equal (P. II.) : heuce, the poly- 
edron Badc-g is a right prism (D. 2, 4), as are also the 
polyedrons Bad-h aud Bcd-h. 

Place the triangle Feh upon Bad, so that F shall 
coincide with B, e with a, and h with d ; then, 
because eE, hH, are perpendicular to the plane Feh, and 
a A, dD, to the plane Bad, /the line eE will take the 
direction aA, and the line hH the direction dD. The 
lines AE and ae are equal, because each is equal to BE 
(B. L, P. XXVIIL). If we take away from the line aE 
the part ae, there will remain the part eE ; and if from 
the Game- line, we take away the part AE, there will re- 
main the part Aa : hence, eE and aA are equal (A. 3) ; 
for a like reason hH is equal to dD : hence, the point 
E will coincide with A, and the point H with D, and 
consequently, the polyedrons Eeli-H and Bad-D will 
coincide throughout, and are therefore equal. 

If from the polyedron Bad -II, we take away the 
part Bad-B, there will remain the prism BAD-H ; 
and if from the same polyedron we take away the part 
Feh-H, there will remain the prism Bad-h : hence, 
these prisms are equal in volume. In like manner, it may 
be shown that the prisms BCD-H and Bcd-h are equal 
in volume. 

The prisms Bad-h, and Bcd-h, have equal bases, be- 
cause these bases are halves of equal parallelograms (B. I., 
P. XXVIIL, C. 1) ; they have also equal altitudes ; they are 
therefore equal (P. V, C.) : hence, the prisms BAD-H and 
BCD-H arc equal (A. 1) ; which was to be proved. 

Cor. Any triangular prism ABD-II, is equal to half of 
tile parallelopipedon AG, which has the same triedral angle 
A, and the same edges AB, AD, and AE. 



190 



GEOMETRY. 



PROPOSITION VIII. THEOREM. 

If two parallelopipedons have a common lower base, and 
their upper bases between the same parallels, they are 
equal in volume. 

Let the parallelopipedons AG and AL have the com- 
mon lower base ABCD, and their upper bases EFGH 
and IKL3I, between the same parallels EK and HL : 
then will they be equal in volume. 

For, the lines EF and 
IK are equal, because each 
is equal to AB ; hence, 
the sum of EF and FI, 
or EI, is equal to the 
sum of FI and IK, or 
FK In the triangular 
prisms AEI-M and 

BFK-L, we have the line AE equal and parallel to 
BF, and EI equal to FK ; hence, the face AEI is 
equal to BFK. In the faces EIMH and FKLG, we have, 
5^= 6^ i?/=:i^ and HEI=GFK: hence, the two faces 
are equal (Bk. I. P. xxviii. C. 3) : the faces ^^2) and W00 
are also equal (P. ' VX) : nence, the prisms are equal (P. 

Y) 

If from the polyedron ABKE-H, we take away the 

prism BFK-L, there will remain the parallelopipedon A G ; 
and if from the same polyedron we take away the prism 
AEI-M, there will remain the parallelopipedon AL : hence, 
these parallelopipedons are equal in volume (A. 3) ; which 
was to be proved. 




BOOK VII. 



191 



PROPOSITION IX. THEOREM. 

If two parallelopipedons have a common loicer base and the 
same altitude, they will be equal in volume. 

Let the parallelopipedons AG. and AL have the com- 
mon lower "base Alt CD and the same altitude; then will 
they be equal in volume. 

Because they have the same altitude, their upper bases 
will lie in the same plane. 
Let the sides I3T and KL 
be prolonged, and also the 
sides FE and GH ; these 
prolongations will form a 
parallelogram Q, which 
will be equal to the com- 
mon base of the given par- 
allelopipedons, because its 
sides are respectively parallel 
and equal to the correspond- 
ing sides of that base. 

Now, if a third parallelopipedon be constructed, having 
for its lower base the parallelogram ABC J), and for its 
upper base JSFOPQ, this third parallelopipedon will be equal 
in volume to the parallelopipedon AG, since they have the 
same lower base, and their upper bases between the same 
parallels, QG, N~~F (P. VIII.). For a like reason, this 
third parallelopipedon will also be equal in volume to the 
parallelopipedon AL : hence, the two parallelopipedons AG 
AL, are equal in volume ; which was to be proved. 

Cor. Any oblique parallelopipedon is equal in volume to 
a right parallelopipedon, having the same base and an equal 
altitude. 




192 



GEOMETRY 



MQ 



LP 



I 




ul 



PROPOSITION X. PROBLEM. 

To construct a rectangular parallelopipedon which shall be 
equal in volume to a right parallelopipedon whose base 
is any parallelogram. 

Let ABCJD-M be a right parallelopipedon, having for 
its base the parallelogram ABGD. 

Through the edges AT and BK pass 
the planes A Q and BP, respectively 
perpendicular to the plane AK, the for- 
mer meeting the face JDL in OQ, and 
the latter meeting that face produced in 
JSTP: then will the polyedron AP be a 
rectangular parallelopipedon equal to the 
given parallelopipedon. It will be a rect- 
angular parallelopipedon, because all of its 
faces are rectangles, and it will be equal to the given 
parallelopipedon, because the two may be regarded as having 
the common base AK (P. VI., C. 1), and an equal altitude 
A (P. IX.). 



Ap 



w 



B 



Cor. 1. A right parallelopipedon, whose base is any paral- 
lelogram, is equal in volume to a rectangular parallelopipedon 
having an equal base and the same altitude. For, the base 
AN is equal to the base A G (B. IV., P. I.) ; and the 
altitude AI is common. 

Cor. 2. An oblique parallelopipedon is equal in volume to 
a rectangular parallelopipedon, having an equal base and an 
equal altitude. 

Cor. 3. Any two parallelopipedons are equal in volume, 
when they have equal bases and equal altitudes. 



BOOK VII. 



193 



PROPOSITION XI. THEOREM. 

Two rectangular parallelopipedons having a common lower 
base, are to each other as their altitudes. 

Let the parallelopipedons AG and AL have the com 
mon lower base ABCJD: then will they be to each other 
as their altitudes AE and AL 

1°. Let the altitudes be commensurable, and suppose, for 
example, that AE is to AI, as 15 is to 8. 

Conceive AE to be divided into 15 equal parts, of 
which AI will contain 8 ; through the points of division 
let planes be passed parallel to ABCD. These planes will 
divide the parallelopipedon AG into 15 parallelopipedons, 
which have equal bases (P. II. C.) and equal altitudes ; 
hence, they are equal (P. X., Cor. 3). 

ISTow, AG contains 15, and AL 8 
of these equal parallelopipedons ; hence, 
AG is to AL, as 15 is to 8, or as 
AE is to AI. In like manner, it may 
be shown that AG is to AL, as AE 
is to AL, when the altitudes are to each 
other as any other whole numbers. 



2°. 
able. 



Let the altitudes be incommensur- 




Now, if AG is not to AL, as AE is to AI, let us 
suppose that, 

AG : AL : : AE : AO, 



in which AO is greater than AI. 

Divide AE into equal parts, such that each shall be 
less than 01 ; there will be at least one point of division 

13 



194 



GEOMETRY. 



m, between and I. Let P denote the parallelopipe- 
don, whose base is ABCB, and altitude Am ; since the 
altitudes AE, Am, are to each other as two whole num- 
bers, we have, 



AG : P : : AE : Am. 

But, by hypothesis, we have, 

AG : AL : : AE : AO \ 
therefore (B. K, P. IV., C), 

AL : P : : AO : Am. 



i 




But AO is greater than Am ; hence, if 
the proportion is true, AL must be greater than P. On 
the contrary, it is less ; consequently, the fourth term of 
the proportion cannot be greater than AL In like manner, 
it may be shown that the fourth term cannot be less than 
AL ; it is, therefore, equal to AL In this case, therefore, 
AG is to AL, as AE is to AL 

Hence, in all cases, the given parallelopipedons are to 
each other as their altitudes ; which was to be proved. 

Scholium. Any two rectangular parallelopipedons whose 
bases are equal, are to each other as their altitudes. 



PROPOSITION Xn. THEOREM. 

Two rectangular parallelopipedons having equal altitudes, are 

to each other as their bases. 

Let the rectangular parallelopipedons AG and ALT have 
the same altitude AE : then will they be to each other aa 
their bases. 



BOOK VII 



195 



For, place them as shown in the figure, and produce the 
plane of the face iVZ, until 
it intersects the plane of the 
face .27(7, in JP Q ; we shall 
thus form a third rectangular 
parallelopipedon A Q. 

The parallelopipedons AG 
and A Q have a common 
base AH ; they are there- 
fore to each other as their 
altitudes AB and AO 
(P. XI.) : hence, we have 
the proportion, 

vol. A G : vol. A Q 




The parallelopipedons AQ and AK have the common base 
AD ; they are therefore to each other as their altitudes 
AD and AM : hence, 



vol. AQ : vol. AK 



AD 



AM. 



Multiplying these proportions, term by term (B. II., P. XII.), 
and omitting the common factor, vol. A Q, we have, 

vol. AG : vol. AK : : AD x AD : AO x AM. 

But AD x AD is equal to the area of the base AD CD : 
and AO X AM is equal to the area of the base AMNO : 
hence, two rectangular parallelopipedons having equal alti- 
tudes, are to each other as their bases ; which was to be 
proved. 



196 



GEOMETRY. 



PROPOSITION XIII. THEOREM. 

Any tico rectangular paraUelopip&fons are to each other as 
tJie products of their bases and altitudes ; that is, as the 
products of their three dimensions. 

* 

Let AZ and AG be 
any two rectangular paral- 
lelopipeclons : then will they 
be to each other as the 
products of their three di- 
mensions. 

For, place them as in the 
figure, and produce the faces 
necessary to complete the 
rectangular parallelopipedon 
AK. The parallelopipedons 
AZ and AK have a com- 
mon base AJST ; hence (P. XL), 

vol AZ : vol AK : : AX : AK 




The parallelopipedons AK and AG have a common 
altitude AK ; hence (P. XII.), 



vol AK : vol AG 



AMNO 



ABCD. 



Multiplying these proportions, term by term, and omitting 
the common factor, vol AK, we have, 

vol AZ : vol AG : : AMNO x AX : ABCD x AK; 

or, since AMKO is equal to AM x AO, and AB CD to 
AB x AD, 

volAZ : vol AG : : A3IxA0xAX: ABxADxAK) 

which was to be proved. 



BOOK VII. 197 

Cor. 1. If we make the three edges A3I, AG, and 

AX, each equal to the linear unit, the parallelopipedon AZ 

will be a cube constructed on that unit, as an edge ; and 

consequently, it will be the unit of volume. Under this 
supposition, the last proportion becomes, 

1 : vol. AG : : 1 : AB X AD x AE ; 

whence, 

vol. AG = AB x AD x XK 

Hence, ffo volume of any rectangular parallelopipedon is 
equal to the product of its three dimensions ; that is, the 
number of times which it contains the unit of volume, is 
equal to the number of linear units in its length, by the 
number of linear units in its breadth, by the number of 
linear units in its height. 

Cor. 2. The volume of a rectangular parallelopipedon i$ 
equal to the product of its base and altitude ; that is, the 
number of times which it contains the unit of volume, is 
equal to the number of superficial units in its base, multi- 
plied by the number of linear units in its altitude. 

Cor. 3. The volume of any parallelopipedon is equal to 
the product of its base and altitude (P. X., C. 2). 

PROPOSITION XIV. THEOREM. 

The volume of any prism is equal to the product of Us 

base and altitude. 

Let ABCDE-K be any prism : then is its volume 
equal to the product of its base and altitude. 

For, through any lateral edge, as AF, pass the planea 
All, AI, dividing it into triangular prisms. These prisms 
will all have a common altitude equal to that of the given 
prism. 



198 



GEOMETRY. 



Now, the volume of any one of the triangular prisms, as 
ABC-IT, is equal to half that of a parallelopipedon con- 
structed on the edges BA, BC, BG 
(P. VTL, C.) ; but the volume of this par- 
allelopipedon is equal to the product of its 
base and altitude (P. XIII., C. 3) ; and 
because the base of the prism is half 
that of the parallelopipedon, the volume 
of the prism is also equal to the pro- 
duct of its base and altitude : hence, 
the sum of the triangular prisms, which 

make up the given prism, is equal to the sum of their 
bases, which make up the base of the given prism, into 
their common altitude ; which was to be proved. 

Cor. Any two prisms are to each other as the products 
of their bases and altitudes. Prisms having equal bases are 
to each other as their altitudes. Prisms having equal alti- 
tudes are to each other as their bases. 




PROPOSITION XV. THEOREM. 



Two triangular pyramids having equal bases and equal alti- 
tudes, are equal in volume. 

Let S-ABC, and S-abc, be two pyramids having their 
equal bases ABC and abc in the same plane, and let AT 
be their common altitude : then will they be equal in vol- 
ume. 

For, if they are not equal in volume, suppose one of 
them, as S-ABC, to be the greater, and let their differ- 
ence be equal to a prism whose base is ABC, and whoso 
altitude is Aa. 



BOOK VII 



199 



Divide the altitude AT into equal parts Ax, xy, &c, 
each of which is less than Act, and let h denote one of 
these parts ; through the points of division pass planes par- 
allel to the plane of the bases ; the sections of the two 
pyramids, by each of these planes, will be equal, namely, 
DEF to def y GHI to ghi, &c. (P. HI., C. 2). 



u - 





On the triangles ABC, DEF, &c, taken as lower bases, 
construct exterior prisms whose edges shall be parallel to 
AS, and whose altitudes shall be equal to h : and on the 
triangles clef, ghi, &c., taken as upper bases, construct 
interior prisms, whose edges shall be parallel to Sa, and 
whose altitudes shall be equal to k. It is evident that the 
sum of the exterior prisms is greater than the pyramid 
S-ABC, and also that the sum of the interior prisms is less 
than the pyramid S~abc : hence, the difference between the 
sum of the exterior and the sum of the interior prisms, is 
greater than the di (Terence between the two pyramids. 

Now, beginning at the bases, the second exterior 
prism EFD-G, is equal to the first interior prism cfd-a, 



200 



GEOMETRY. 



because they hava the same altitude To, and their bases 
EFD, efd, are equal: for a like reason, tie third exterior 
prism IIIG-K, and the second interior prism hig-d, are 
equal, and so on to the last in each set : hence, each of the 
exterior prisms, excepting the first BCA-D, has an equal 
corresponding interior prism ; the prism B CA-D, is, there- 
fore, the difference between the sum of all the exterior 
prisms, and the sum of all the interior prisms. But the 
difference between these two sets of prisms is greater than 
that between the two pyramids, which latter difference was 
supposed to be equal to a prism whose base is BCA, and 
whose altitude is equal to Aa, greater than h ; conse- 
quently, the prism BCA-B is greater than a prism having 
the same base and a greater altitude, which is impossible . 
hence, the supposed inequality between the two pyramids 
cannot exist ; they are, therefore, equal in yolume ; which 
was to be proved. 



PROPOSITION XVI. THEOREM. 

Any triangular prism may be divided into three triangular 
pyramids, equal to each other in volume. 



Let ABC-B be a triangular 
prism : then can it be divided into 
three equal triangular pyramids. 

For, through the edge A C> 
pass the plane ACB] and through 
the edge EF pass the plane 
EFC. The pyramids ACE-F and 
ECB-F, have their bases ACE 
and ECD equal, because they are 
halves of the same parallelogram 
ACBE\ and they have a common 




BOOK VII. 



201 



altitude, because their bases are in the same plane AD, and 
their vertices at the same point F \ hence, they are equal 
in volume (P. XT.)- The pyramids ABC-F and DEF-C, 
have their bases ABC and JDEF, equal because they are 
the bases of the given prism, and their altitudes are equal 
because each is equal to the altitude of the prism ; they 
are, therefore, equal in volume : hence, the three pyramids 
.into which the prism is divided, are all equal in volume ; 
which was to he proved. 

Cor. 1." A triangular pyramid is one-third of a prism, 
having an equal base and an equal altitude. 

Cor. ' 2. The volume of a triangular pyramid is equal to 
one-third of the product of its base and altitude. 



PROPOSITION XVII. THEOREM. 

* 

The volume of any pyramid is equal to one-third of the 
product of its base and altitude. 

Let S-ABCDE, be any pyramid: then is its volume 
equal to one-third of the product of its base and altitude. 

For, through any lateral edge, as BE, 
pass the planes SEB, SEC, dividing the 
pyramid into triangular pyramids. The alti- 
tudes of these pyramids will be equal to 
each other, because each is equal to that 
of the given pyramid. Now, the volume 
of each triangular pyramid is equal to one- 
third of the product of its base and alti- 
tude (P. XVI, C. 2) ; hence, the sum of 
the volumes of the triangular pyramids, is 
equal to one-third of the product of the sum of their basea 




202 GEOMETRY. 

by their common altitude. But the sum of the triangular 
pyramids is equal to the given pyramid, and the sum of 
their bases is equal to the base of the given pyramid : 
hence, the volume of the given pyramid is equal to one- 
third of the product of its base and altitude ; which was to 
be proved. 

Cor. 1. The volume of a pyramid is equal to one-third 
of the volume of a prism having an equal base and an equal 
altitude. 

Cot. 2. Any two pyramids are to each other as the 
products of their bases and altitudes. Pyramids having equal 
bases are to each other as their altitudes. Pyramids having 
equal altitudes are to each other as their bases. 

Scholium. The volume of a polyedron may be found by 
dividing it into triangular pyramids, and computing their 
volumes separately. The sum of these volumes will be equal 
to the volume of the polyedron. 



PROPOSITION XVIII. THEOREM. 

The volume of a frustum of any triangular pyramid is 
equal to the sum of the volumes of three pyramids 
whose common altitude is that of the frustum, and tohose 
bases are the lower base of the frustum, the tipper base 
of the frustum, and a mean proportional between the two 
bases. 

Let FGH-h be a frustum of any triangular pyramid : 
then will its volume be equal to that of three pyramids 
whose common altitude is that of the frustum, and whose 
bases are the lower base FGJJ, the upper base fgh, and 
a mean proportional between their bases. 



BOOK VII. 



203 




For, through the edge FK, pass the plane FHg, and 
through the edge fg, pass the plane fgH, dividing the 
frustum into three pyramids. The pyra- 
mid g-FGII, has for its base the lower 
base FGH of the frustum, and its al- 
titude is equal to that of the frustum, 
because its vertex g, is in the plane of 
the upper base. The pyramid H-fgh, 
has for its base the upper base fgh cf 
the frustum, and its altitude is equal to 
that of the frustum, because its vertex 
lies in the plane of the lower base. 

The remaining pyramid may be regarded as having the 
triangle FfH for its base, and the point g for its vertex. 
From g, draw gK parallel to fF, and draw also KH and 
Kf. Then will the pyramids K-FfH and g-Ffll, be equal ; 
for they have a common base, and their altitudes are equal, 
because their vertices K and g are in a line parallel to 
the base (B. VI., P. XII., C. 2). 

Xow, the pyramid K-Ffll may be regarded as having 
FKII for its base and / for its vertex. From Jl, draw 
KL parallel to Gil ; it will be parallel to gh : then will 
the triangle FKL be equal to fgh, for the side FK is 
equal to fg, the angle F to the angle /, and the angle K 
to the angle g. But, FKII is a mean proportional between 
FKL and FGII (B. IV., P. XXIV, C), or between fgh 
and JFGII The pyramid f-FKII, has, therefore, for its 
base a mean proj^ortional between the upper and lower bases 
of the frustum, and its altitude is equal to that of the frus- 
tum ; but the pyramid f-FKII is equal in volume to the 
pyramid g-Ffll: hence, the volume of the given frustum is 
equal to that of three pyramids whose common altitude is 
equal to that of the frustum, and whose bases are the upper 
base, the lower base, and a mean proportional between 
them ; which icas to be proved. 



204: 



GEOMETRY. 



Cor. The volume of the frustum of any pyramid is 
equal to the sum of the volumes of three pyramids whose 
common altitude is that of the frustum, and whose bases 
are the lower base of the frustum, the upper base of the 
frustum, and a mean proportional between them. 

For, let ABCDE-e be a frustum of 
any pyramid. Through, any lateral edge, as 
eE, pass the planes eEBb, eECc, divid- 
ing it into triangular frustums. Now, the 
sum of the volumes of the triangular frus- 
tums is equal to the sum of three sets of 
pyramids, whose common altitude is that of 
the given frustum. The bases of the first 
set make up the lower base of the given 
frustum, the bases of the second set make up the upper base 
of the given frustum, and the bases of the third set make 
up a mean proportional between the upper and lower base 
of the given frustum : hence, the sum of the volumes of 
the first set is equal to that of a pyramid whose altitude is 
that of the frustum, and whose base is the lower base of 
of the frustum ; the sum of the volumes of the second set 
is equal to that of a pyramid whose altitude is that of the 
frustum, and whose base is the upper base of the frustum ; 
and, the sum of the third set is equal to that of a pyra- 
mid whose altitude is that of the frustum, and whose base 
is a mean proportional between the two bases. 




PROPOSITION XIX. 



THEOREM. 



Similar triangular prisms are to each other as the cubes of 

their homologous edges. 

Let CBD-P, cbd-p, be two similar triangular prisms, 
and let B C, be, be any two homologous edges : then will 
the prism CBD-P be to the prism cbd-p, as BC 3 to be 3 



BOOK VII. 



205 




For, the homologous angles B and b are equal, and 
the faces which bound them are similar (D. 16) : hence, 
these triedral angles may be 
applied, one to the other, so 
that the angle cbd will coin- 
cide with CBD, the edge ha 
with BA. In this case, the 
prism cbd-p will take the 
position Bcd-p. From A 
draw AH perpendicular to 

the common base of the prisms : then will the plane BAH 
be perpendicular to the plane of the common base (B. VI., 
P. XVI.). From a, in the plane BAH, draw ah 
perpendicular to BH : then will ah also be perpendicular 
to the base BBC (B. VI., P. XVLI.) ; and AH, ah, will 
be the altitudes of the two prisms. 

Since the bases CBB, cbd, are similar, we have (B. IV., 
P. XXV), 




base CBB 



base cbd 



CH 



cb . 



Now, because of the similar triangles ABH, aBh, and of 
the similar parallelograms AC, ac, we have, 

AH : ah : : CB : cb ; 
hence, multiplying these proportions term by term, we have, 



nr 7?3 



base CBB x AH : base cbd x ah : : CB 



cb . 



But, base CBB x AH is equal to the volume of the prism 
CBB-A, and base cbd X ah is equal to the volume of 
the prism cbd-p ; hence, 



prism CBB-P : prism cbd-p 
which was to be proved. 



CB' 



cb 3 ] 



206 



GEOMETRY. 



Cor. 1. Any two similar prisms are to each other as 
the cubes of their homologous edges. 

For, since the prisms are similar, their bases are similar 
polygons (D. 16) ; and these similar polygons may each be 
divided into the same number of similar triangles, similarly 
placed (B. IV., P. XXVI.) ; therefore, each prism may be 
divided into the same number of triangular prisms, having 
their faces similar and like placed ; consequently, the tri- 
angular prisms are similar (IX 16). But these triangular 
prisms are to each other as the cubes of their homologous 
edges, and being like parts of the polygonal prisms, the 
polygonal prisms themselves are to each other as the cubes 
of their homologous edges. 

Cor. 2. Similar prisms are to each other as the cubes 
of their altitudes, or as the cubes of any other homologous 
lines. 



PROPOSITION XX. THEOREM. 

Similar pyramids are to each other as the cubes of their 

homologous edges. 

Let S-ABCDE, and S-abcde, be two similar pyra- 
mids, so placed that their homologous angles at the vertex 
shall coincide, and let AB and ab be 
any two homologous edges : then will the 
pyramids be to each other as the cubes 
of AB and ab. 

For, the face SAB, being similar to 
Sab, the edge AB is parallel to the 
edge ab, and the face SBC being simi- 
lar to Sbc, the edge BC is parallel to 
be ; hence, the planes of the bases are 
parallel (B. VI., P. XIII.). 




BOOK VII. 



207 



Draw SO perpendicular to the base ABODE ; it will 
also "be perpendicular to the base cibcde. Let it pierce that 
plane at the point o : then will SO 
be to So, as SA is to Sa (P. IH.)> S 

or as AB is to ab ; hence, 



\SO : \So : : ^S : a&. 



But the bases being similar polygons, we 
have (B. IV., P. XXVH), 



base AB CDE : base abcde 



AB' 




Multiplying these proportions, term by term, we have, 



base ABODE x \SO : base abcde x \ So :: AB : ab . 

But, base ABODE x ££6> is equal to the volume of tho 
pyramid S- ABODE, and base abcde X %So is equal to 
the volume of the pyramid S-abcde ; hence, 

pyramid S-AB ODE : pyramid S-abcde : : AB • ab ; 
which was to be proved. 



Cor. Similar pyramids are to each other as the cubes of 
their altitudes, or as tho cubes of any other homologous 
lines. 



208 GEOMETRY. 



GENEEAL FOEMTJLAS. 

If we denote the volume of any prism by V, its base 
by B, and its altitude by JS, we shall have (P. XIV.) , 

F = B x H (1.) 

If we denote the volume of any pyramid by V, its 
base by B, and its altitude by JI, we have (P. XVII.), 

V= fB X H (2.) 

If we denote the volume of the frustum of any pyramid 
by V, its lower base by B, its upper base by b, and 
its altitude by M, we shall have (P. XVHL, C), 

V = i(B + b + VB X b) X M • • (3.) 



EEGULAE POLYEDEONS. 

A Regular Polyedeon is one whose faces are all equal 
regular polygons. 

There are five regular polyedrons, namely : 

1. The Teteaedeon, or regular pyramid — a polyedron 
bounded by four equal equilateral triangles. 

2. The Hexaedeon, or cube— a polyedron bounded by 
gix equal squares. 

3. The Octaedeon — a polyedron bounded by eight equal 
equilateral triangles. 

4. The Dodecaedeon — a polyedron bounded by twelve 
equal and regular pentagons. 



BOOK VII. 209 

5. The Icosaedkox — a polyedron bounded by twenty 
equal equilateral triangles. 

In the Tetraedron, the triangles are grouped about the 
polyedral angles in sets of three, in the Octaedron they are 
grouped in sets of four, and in the Icosaedron they are 
grouped in sets of five. JSToav, a greater number of equi- 
lateral triangles cannot be grouped so as to form a salient 
polyedral angle ; for, if they could, the sum of the plane 
angles formed by the edges would be equal to, or greater 
than, four right angles, which is impossible (B. VI., P. XX.). 

In the Hexaedron, the squares are grouped about the 
polyedral angles in sets of three. Now, a greater number 
of squares cannot be grouped so as to form a salient polye- 
dral angle ; for the same reason as before. 

In the Dodecaedron, the regular pentagons are grouped 
about the polyedral angles in sets of three, and for the same 
reason as before, they cannot be grouped in any greater 
number, so as to form a salient polyedral angle. 

Furthermore, no other regular polygons can be grouped 
so as to form a salient polyedral angle ; therefore, 

Only jive regular polyedrons can be formed. 

14 



BOOK VIII. 



THE CTLINDEE, THE CONE, AND THE SPHERE 



DEFINITIONS. 



1. A Cylinder is a volume which may be generated by 
a rectangle revolving about one of its sides as an axis. 

Thus, if the rectangle ABCD be turned about the side 
AB, as an axis, it will generate the cylinder FGtCQ-P. 

The fixed line AB is called the axis 
of the cylinder ; the curved surface generated 
by the side CD, opposite the axis, is called 
the convex surface of the cylinder ; the equal 
circles FGCQ, and EHDP, generated by 
the remaining sides BC and AB, are called 
bases of the cylinder ; and the perpendicular 
distance between the planes of the bases, is 
called the altitude of the cylinder. 

The line BC, which generates the convex surface, is, in 
any position, called an element of the surface ; the elements 
are all perpendicular to the planes of the bases, and any 
one of them is equal to the altitude of the cylinder. 

Any line of the generating rectangle ABCB, as IK, 
which is perpendicular to the axis, will generate a circle 
whose plane is perpendicular to the axis, and which is equa 
to either base : hence, any section of a cylinder by a plan 
perpendicular to the axis, is a circle equal to either base. 
Any section, FCJDJE, made by a plane through the axis 
i&s a rectangle double the generating rectangle. 




BOOK VIII. 



211 



2. S tattt.a-r . Cylindees a^e those which may be generated 
by similar rectangles revolving about homologous sides. 

The axes of similar cylinders are proportional to the radii 
of their bases (B. IV., D. 1) ; they are also proportional to 
any other homologous lines of the cylinders. 




3. A prism is said to be inscribed 
in a cylinder, when its bases are in- 
scribed in the bases of the cylinder. 
In this case, the cylinder is said to 
be circumscribed about the prism. 

The lateral edges of the inscribed 
prism are elements of the surface of 
the circumscribing cylinder. 



4. A prism is said to be circum- 
scribed about a cylinder, when its 

bases are circumscribed about the bases of the cylinder. 
In this case, the cylinder is said to be inscribed in the 
prism. 

The lines which join the corres- 
ponding points of contact in the upper 
and lower bases, are common to the 
surface of the cylinder and to the 
lateral faces of the prism, and they 
are the only lines which are common. 
The lateral faces of the prism are said 
to be tangent to the cylinder along 
these lines, which are then called de- 
ments of contact. 

5. A Cone is a volume which may be generated by a 
right-angled triangle revolving about one of the sides adja- 
cent to the right angle, as an axis. 





212 GEOMETRY. 

Thus, if the triangle SAB, right-angled at A, be turned 
about the side SA, as an axis, it will generate the cone 
S- CD-BE. 

The fixed line SA, is called the 
axis of the cone ; the curved surface 
generated by the hypothenuse SB, is * 
called the convex surface of the cone ; 
the circle generated by the side AB, 
is called the base of the cone / and 
the point S, is called the vertex of 
the cone ; the distance from the vertex 
to any point in the circumference of the 

base, is called the slant height of the cone ; and the per- 
pendicular distance from the vertex to the plane of the base, 
is called the altitude of the cone. 

The line SB, which generates the convex surface, is, in 
any position, called an element of the surface / the elements 
are all equal, and any one is equal to the slant height ; the 
axis is equal to the altitude. 

Any line of the generating triangle SAB, as GH, 
which is perpendicular to the axis, generates a circle whose 
plane is perpendicular to the axis : hence, any section of a 
cone by a plane perpendicular to the axis, is a circle. Any 
section SBC, made by a plane through the axis, is an 
isosceles triangle, double the generating triangle. 

6. A "Truncated Cone is that portion of a cone included 
between the base and any plane which cuts the cone. 

When the cutting plane is parallel to the plane of the 
base, the truncated cone is called a Frustum of a Cone, and 
the intersection of the cutting plane with the cone is called 
the upper base of the frustum ; the base of the cone is 
called the lower base of the frustum. 



BOOK VIII. 



21a 




If the trapezoid HGAJB, right-an- 
gled A and G, be revolved about 
AG, as an axis, it will generate a frus- 
tum of a cone, whose bases are ECJDB 
and FKJS, whose altitude is AG, and 
whose slant height is JBH. 



1. Similar Coxes are those which may be generated 
by similar right-angled triangles revolving about homologous 

sides. 

The axes of similar cones are proportional to the radii 
of their bases (B. IY., D. 1) ; they are also proportional to 
any other homologous lines of the cones. 



8. A pyramid is said to be in- 
scribed in a cone, when its base is 
inscribed in the base of the cone, and 
when its vertex coincides with that of 
the cone. 

The lateral edges of the inscribed 
pyramid are elements of the surface of 
the circumscribing cone. 




9. A pyramid is said to be circumscribed about a cone, 
when its base is circumscribed about the base of the cone, 
and when its vertex coincides with that of the cone. 
# In this case, the cone is said to be inscribed in the 
pyramid. 

The lateral faces of the circumscribing pyramid are tan- 
gent to the surface of the inscribed cone, along lines which 
are called elements of contact. 



10, A frustum of a pyramid is inscribed in a frustum 



214 GEOMETRY. 

of a co?ie, when its bases are inscribed in the bases of the 
frustum of the cone. 

The lateral edges of the inscribed frustum of a pyramid 
are elements of the surface of the circumscribing frustum of 
a cone. 

• 

11. A frustum of a pyramid is circumscribed about a 
frustnm of a cone, when its bases are circumscribed about 
those of the frustum of the cone. 

Its lateral faces are tangent to the surface of the frustum 
of the cone, along lines which are called elements of contact. 

12. A Spheee is a volume bounded by a surface, every 
point of which is equally distant from a point within called 

the centre. 

A sphere may be generated by a semicircle revolving 
about its diameter as an axis. 

13. A Radius of a sphere is a straight line drawn from 
the centre to any point of the surface. A Diameter is any 
straight line drawn through the centre and limited at both 
extremities by the surface. 

All the radii of a sphere are equal : the diameters are 
also equal, and each is double the radius. 

14. A Spherical Sector is a volume which may be gen- 
erated by a sector of a circle revolving about a diameter of 
the circle lying without it. 

The surface generated by the arc is called the base of 
the sector. 

15. A plane is Tangent to a Sphere when it touches 
it in a single point. 

10. A Zone is a portion of the surf-ice of a sphere 
included between two parallel planes. The bounding lines 



BOOK VIII. 



215 



of the sec? ons are called bases of the zone, and the distance 
between tr,3 planes is called the altitude of the zone. 

If one of the planes is tangent to the sphere, the zone 
has hut one base. 

17. A Spherical Segment is a portion of a sphere in- 
cluded between two parallel planes. The sections made by 
the planes are called bases of the segment, and the distance 
between them is called the altitude of the segment. 

If one of the planes is tangent to the sphere, the seg- 
ment has but one base. 

The Cylinder, the Cone, and the Sphere, are sometimes 
called The Three Round Bodies. 



PROPOSITION I. THEOREM. 



The convex surface of a cylinder is equal to the circum- 
ference of its base multiplied by the altitude. 

Let ABD be the base of a cylinder whose altitude is 
IT : then will its convex surface be equal to the circum- 
ference of its base multiplied by the altitude. 

For, inscribe within the cylinder a 
prism Avhose base is a regular polygon. 
The convex surfice of this prism will 
be equal to the perimeter of its base 
multiplied by its altitude (B. VII., P. I.), 
whatever may be the number of sides 
of its base. But, when the number of 
sides is infinite (B. V., P. X., C. 1), the 
convex surface of the prism coincides with 
that of the cylinder, the perimeter of 




216 



GEOMETRY. 



the base of the prism coincides with the circumference of 
the base of the cylinder, and the altitude of the prism is 
the same as that of the cylinder : hence, the convex surface 
of the cylinder is equal to the circumference of its base 
multiplied by the altitude ; which was to be proved. 

Cor. The convex surfaces of cylinders having equal alti- 
tudes are to each other as the circumferences of their bases. 



PEOPOSITION II. THEOEEM. 

The volume of a cylinder is equal to the product of its 

base and altitude. 

Let ABJD be the base of a cylinder whose altitude is 
H\ then will its volume be equal to the product of its 
base and altitude. 

For, inscribe within it a prism whose 
base is a regular polygon. The volume 
of this prism is equal to the product 
of its base and altitude (B. VII., P. 
XIV.), whatever may be the number of 
sides of its base. But, when the num- 
ber of sides is infinite, the prism coin- 
cides with the cylinder, the base of the 
prism with the base of the cylinder, and 
the altitude of the prism is the same 

as that of the cylinder : hence, the volume of the cylinder 
is equal to the product of its base and altitude ; which was 
to be proved. 

Cor. 1. Cylinders are to each other as the products of 
their bases and altitudes ; cylinders having equal bases are 
to each other as their altitudes ; cylinders having equal alti- 
tudes are to each other as their bases. 




BOOK VIII. 



217 



Cor. 2. Similar cylinders are to each other as the cubes 
of their altitudes, or as the cubes of the radii of their 
bases. 

For, the bases are as the squares of their radii (B. V., 
P. Xm.), and the cylinders being similar, these radii are to 
each other as their altitudes (D. 2) : hence, the bases are 
as the squares of the altitudes ; therefore, the bases multiplied 
by the altitudes, or the cylinders themselves, are as the 
cubes of the altitudes. 



PROPOSITION III. THEOREM. 

The convex surface of a cone is equal to the circumference 
of its base multiplied by half the slant height. 

Let S-ACD be a cone whose base is A CD, and whose 
slant height is SA : then will its convex surface be equal 
to the circumference of its base multiplied by half the slant 
height. 

For, inscribe within it a right pyramid. 
The convex surface of this pyramid is 
equal to the perimeter of its base mul- 
tiplied by half the slant height (B. VII., 
P. IV.), whatever may be the number 
of sides of its base. But when the num- 
ber of sides of the base is infinite, the 
convex surface coincides with that of the 
cone, the perimeter of the base of the pyramid coincides with 
the circumference of the base of the cone, and the slant height 
of the pyramid is equal to the slant height of the cone : 
hence, the convex surface of the cone is equal to the cir- 
cumference of its base multiplied by half the slant height; 
which was to be proved. 




218 



GEOMETRY. 



PROPOSITION IV. THEOREM. 

The convex surface of a frustum of a cone is equal to 
half the sum of the circumferences of its two bases 
multiplied by the slant height. 

Let BIA-B be a frustum of a coue, BIA and EGB 
its two bases, and EB its slant height: then is its convex 
surface equal to half the sum of the circumferences of its 
two bases multiplied by its slant height. 

For, inscribe within it the frustum 
of a right pyramid. The convex sur- 
face of this frustum is equal to half 
the sum of the perimeters of its bases, 
multiplied by the slant height (B. VII., 
P. IV., C), whatever may be the 
number of its lateral faces. But when 
the number of these faces is infinite, 

the convex surface of the frustum of the pyramid coincides 
with that of the cone, the perimeters of its bases coincide 
with the circumferences of the bases of the frustum of the 
cone, and its slant height is equal to that of the cone: 
hence, the convex surface of the frustum of a cone is equal 
to half the sum of the circumferences of its bases multiplied 
by the slant height ; which was to be proved. 

Scholium. From the extremities A and Z>, and from 
the middle point I, of a line AB, let the lines AO> BC, 
and IK, be drawn perpendicular to a line OC: then will 
IK be equal to half the sum of AO and BC. For, 
draw Bd and 1% perpendicular to AO : then, because Al 
is equal to IB, we shall have Ai equal to id (B. IV., P. 
XV.), and consequently to Is ; that is, A exceeds IK 




BOOK VIII. 



219 



as much as IK exceeds D C : hence, IK is equal to the 
half sum of AO and DC. 

Now, if the line AD be revolved about 0(7, as an 
axis, it will generate the surface of a frustum of a cone^ 
whose slant height is AD ; the point I will generate a 
circumference which is equal to half the sum of the circum- 
ferences generated by A and D : hence, if a straight line 
be revolved about another straight Ime, it will generate a 
surface whose measure is equal to the product of the gene- 
rating line and the circumference generated by its middle 
point. 

This proposition holds true when the line AD meets 
0C y and' also when AD is parallel to OC. 



PROPOSITION V. THEOREM. 

The volume of a cone is equal to its base multiplied by 

one-third of its altitude. 

Let ABDE be the base of a cone whose vertex is J3, 
and whose altitude is JSo : then will its volume be equal to 
the base multiplied by one-third of the altitude. 

For, inscribe in the cone a right 
pyramid. The volume of this pyramid 
is equal to its base multiplied by one- 
third of its altitude (B. VII, P. XVIL), 
whatever may be the number of its 
lateral faces. But, when the number 
of lateral faces is infinite, the pyramid 
coincides with the cone, the base of 
the pyramid coincides with that of the 

cone, and their altitudes are equal : hence, the volume of a 
cone is equal to the base multiplied by one-third of the 
altitude ; which was to be proved. 







220 



GEOMETRY. 



Cor. 1. A cone is equal to one-third of a cylinder hav- 
ing an equal base and an equal altitude. 

Cor. 2. Cones are to each other as the products of 
their bases and altitudes. Cones having equal bases are to 
each other as their altitudes. Cones having equal altitudes 
are to each other as their bases. 



PROPOSITION YI. THEOREM. 

The volume of a frustum of a cone is equal to the sum 
of the volumes of three cones, having for a common 
altitude the altitude of the frustum, and for bases the 
loicer base of the frustum, the upper base of the frus 
turn, and a mean proportional between the bases. 

Let BIA be the lower base of a frustum of a cone, 
JEGD its upper base, and C its altitude : then will its 
volume be equal to the sum of three cones whose common 
altitude is OC, and whose bases are the lower base, the 
upper base, and a mean proportional between them. 

For, inscribe a frustum of a right 
pyramid in the given frustum. The 
volume of this frustum is equal to 
the sum of the volumes of three 
pyramids whose common altitude is 
that of the frustum, and whose bases 
are the lower base, the upper base, 
and a mean proportional between the 
two (B. VIL, P. XVIII.), whatever 

may be the number of lateral faces. But when the numbei 
of faces is infinite, the frustum of the pyramid coincides 
with the frustum of the cone, its bases with the bases of 
the cone, the three pyramids become cones, and their altitudes 







BOOK VIII. 221 

are equal to that of the frustum ; hence, the volume of the 
frustum of a cone is equal to the sum of the volumes of 
three cones whose common altitude is that of the frustum, 
and whose bases are the lower base of the frustum, the 
tipper base of the frustum, and a mean proportional between 
them ; which was to be proved. 



PEOPOSITION VII. THEOEEM. 
Any section of a sphere made by a plane, is a circle. 

Let C be the centre of a sphere, CA one of its 
radii, and AMB any section made by a plane : then will 
this section be a circle. 

For, draw a radius CO perpen- 
dicular to the cutting plane, and let 
it pierce the plane of the section at 
0. Draw radii of the sphere to any 
two points M, M', of the curve which 
bounds the section, and join these 
points with : then, because the radii 
CM, CM' are equal, the points 

M, M', will be equally distant from (B. VI., P. V., C.) ; 
hence, the section is a circle ; ichich was to be proved. 

Cor. 1. When the cutting plane passes through the centre 
of the sphere, the radius of the section is equal to that of 
the sphere ; when the cutting plane does not pass through 
the centre of the sphere, the radius of the section will bo 
less than that of the sphere. 

A section whose plane passes through the centre of the 
Bphere, is called a great circle of the sphere. A section 
whose plane does not pass through the centre of the sphere, 




222 GEOMETRY. 

is called a small circle of the sphere. All great circles of 
the same, or of equal spheres, are equal. 

Cor. 2. Any great circle divides the sphere, and also 
the surface of the sphere, into equal parts. For, the parts 
may he so placed as to coincide, otherwise there would be 
some points of the surface unequally distant from the centre, 
which is impossible. 

Cor. 3. The centre of a sphere, and the centre of any 
small circle of that sphere, are in a straight line perpen- 
dicular to the plane of the circle. 

Cor. 4. The square of the radius of any small circle is 
equal to the square of the radius of the sphere diminished 
by the square of the distance from the centre of the sphere 
to the plane of the circle (B. IV., P. XI., C. 1) : hence, 
circles which are equally distant from the centre, are equal ; 
and of two circles which are unequally distant from the 
centre, that one is the less whose plane is at the greater 
distance from the centre. 

Cor. 5. The circumference of a great circle may always 
be made to pass through any two points on the surface of 
a sphere. For, a plane can always be passed through these 
points and the centre of the sphere (B. VL, P. II.), and its 
section will be a great circle. If the two points are the 
extremities of a diameter, an infinite number of planes can 
be passed through them and the centre of the sphere (B. VI., 
P. I., S.) ; in this case, an infinite number of great circles 
can be made to pass through the two points. 

Cor. 6. The bases of a zone are the circumferences of 
circles (D. 16), and the bases of a segment of a sphere are 
circles. 



BOOK \III. 



223 




PROPOSITION Till. THEOEEM. 

Any plane perpendicular to a radius of a sphere at its 
extremity, is tangent to the sphere at that point. 

Let C be the centre of a sphere, CA any radius, and 
FAG a plane perpendicular to CA at A : then will the 
plane FAG- be tangent to the sphere at A. 

For, from any other point of the 
plane, as 31, draw the line M G : 
then because CA is a perpendicular 
to the plane, and CJI an oblique 
line, CJI Trill be greater than CA 
(B. VI., P. V.) : hence, the point 31 
lies without the sphere. The plane 
FAG, therefore, touches the sphere 

at A, and consequently is tangent to it at that point , 
which was to be proved. 

Scholium. It may be shown, by a course of reasoning 
analogous to that employed in Book ILT., Propositions XI., 
XIX, XIII., and XIV., that two spheres may have any one 
of six positions with respect to each other, viz. : 

1°. "When the distance between their centres is Greater than 
the sum of their radii, they are external, one to the other : 

2°. When the distance is equal to the sum of their 
radii, they are tangent, externally : 

3°. When this distance is less than the sum, and greater 
than the difference of their radii, they intersect each other : 

4°. When this distance is equal to the difference of theii 
radii, they are tangent internally : 

5°. When this distance is less than the difference of their 
radii, one is wholly within the other : 

6°. When this distance is equal to zero, they have a 
common centre, or, are concentric. 



224 GEOMETRY. 



DEFINITIONS. 



1°. If a semi-circumference be divided into equal arcs, the 
chords of these arcs form half of the perimeter of a regular 
inscribed polygon ; this half perimeter is called a regular 
semi-perimeter. The figure bounded by the regular semi- 
perimeter and the diameter of the semi-circumference is called 
a regular semi-polygon. The diameter itself is called the 
axis of the semi-polygon. 

2°. If lines be drawn from the extremi- 
ties of any side, and perpendicular to the 
axis, the intercepted portion of the axis is 
called the projection of that side. 

The broken line APCPGP is a regu- 
lar semi-perimeter ; the figure bounded by J 
it and the diameter AP, is a regular 
semi-polygon, AP is its axis, UK is the 
projection of the side PC, and the axis, 
AP, is the projection of the entire semi-perimeter. 

* PROPOSITION IX. LEMMA. 

If a regular semi-polygon he revolved about its axis, the 

surface generated by the semi-perimeter will be equal to 

the axis multiplied by the circumference of the inscribed 
circle. 

Let ABCDEF be a regular semi-polygon, AF its axis, 
and ON its apothem : then will the surface generated by 
the regular semi-perimeter be equal to AF x circ. ON. 

From the extremities of any side, as PE, draw PI 
and FIT perpendicular to AF ; draw also N3I perpen- 
dicular to AF, and EK perpendicular to PI. Now, the 
sur&ce generated by EP is equal to PE x circ. NJf 



BOOK VIII. 



225 



(P. 3V., S.). But, because the triangles EDK and ONM 
are similar (B. IV., P. XXI.), we have, 

DE : EK or IB : : ON : NM : : circ. ON : circNM; 

whence, 

DE x tire. JVM — IH X circ. ON ; 

that is, the surface generated by any side 
is equal to the projection of that side 
multiplied by the circumference of the in- 
scribed circle : hence, the surface gene- 
rated by the entire semi-perimeter is equal 
to tho'sum of the projections of its sides, 
or the axis, multiplied by the circumfer- 
ence of the inscribed circle ; which was to be proved. 

Cor. The surface generated by any portion of the perim- 
eter, as CDE, is equal to its projection FIT, multiplied 
by the circumference of the inscribed circle. 




PROPOSITION 






THEOPvEM. 



The surface of a sphere is equal to its diameter multiplied 
by the circumference of a great circle. 

Let ABODE be a semi-circumference, 
its centre, and AE its diameter : then 
will the surface of the sphere generated 
by revolving the semi-circumference about 
AE, be equal to AE x circ. OE. 

For, the Semi-circumference may be re- 
garded as a regular semi-perimeter with an 
infinite number of sides, whose axis is AE, 
and the radius of whose inscribed circle 
is OE : hence (P. IX.), the surface generated by it is equal 

to AE x circ. 0E-, which was to be proved. 

15 




226 



GEOMETRY. 



Cor. 1. The circumference of a great circle is equal to 
2«OE (B. Y., P. XVI.) : hence, the area of the surface 
of the sphere is equal to 2 0E X 2*0E, or to 4* OE 2 ; 
that is, £Ae «rea o/ ZAe surface o/ a sphere is equal to four 

great circles. 

Cor. 2. The surface generated by any 
arc of the semicircle, as BC, will be a 
zone, whose altitude is equal to the pro- 
jection of that arc on the diameter. But, 
the arc 5(7 is a portion of a semi- 
perimeter having an infinite number of 
sides, and the radius of whose inscribed 
circle is equal to that of the sphere : 
hence (P. IX., C), the surface of a zone 
is equal to its altitude multiplied by the circumference of a 
great circle of the sphere. 

Cor. 3. Zones, on the same sphere, or on equal spheres, 
are to each other as their altitudes. 




PKOPOSITXON XI. LEMMA. 

If a triangle and a rectangle having the same base and 
equal altitudes, be revolved about the common base, the 
volume generated by the triangle will be one-third of that 
generated by the rectangle. 

Let ABC be a triangle, and EFBC a rectangle, having 
the same base BC, and an equal altitude AD, and let 
them both be revolved about BC: then will the volume 
generated by ABC be one-third of that generated b> 

EFB C. 

For, the cone generated by the right-angled triangl 
ABB, is equal to one-third of the cylinder generated by 



BOOK VIII. 



227 




the rectangle ADDF (P. V., C. 1) ; and the cone generated 
by the triangle ADC, is equal to one-third of the cylinder 
generated by the rectangle AD CF. 
But, when AD falls within the 
triangle, the sum of the cones 
generated by ADD and ADC, 
is equal to the volume generated 
by the triangle AD C ; and the 
sum of the cylinders generated by 

ADDF and ADCF, is equal to the volume generated by 
the rectangle FFDC. "When AD falls without the triangle, 
the difference of the cones generated by ADD and ADO, 
is equal to the volume generated by AD C ; and the differ- 
ence of the cylinders generated by ADDF and ADCF, is 
equal to the volume generated by FFDC: hence, in either 
case, the volume generated by the triangle ADC, is equal 
to one-third of the volume generated by the rectangle 
EFD C ; which teas to be proved. 

Cor. The volume of the cylinder generated by FFDC, 
is equal to the product of its base and altitude, or to 
*2lD 2 x DC : hence, the volume generated by the triangle 
ADC, is equal to 



\*Air x DC. 



PROPOSITION XII. LEMMA. 

If an isosceles triangle he revolved about a straight line 
passing through its vertex, the volume generated will be 
equal to the surface generated by the base multiplied by 
one-third of the altitude. 

Let CAD be an isosceles triangle, C its vertex, AD 
its base, CI its altitude, and let it bo revolved about the 
line CD, as an axis : then will the volume generated be 
equal to surf. AD x J CI. 



228 



GEOMETRY. 



/ 



There may be two cases : the base, or base produced, 
may meet the axis ; or, the base may be parallel to the axis. 

1°. Suppose the base, when 
produced, to meet the axis at 
D ; draw AM, IK, and BJST, 
perpendicular to CD, and BO 
parallel to DC. Now, the 
volume generated by CAD is ^ 
equal to the difference of the 
volumes generated by CAD and CDD ; hence (P. XL, C), 




m !k hsr 



X 



D 



voWAB=^AM 2 x CD-^BJST 2 x CD=^(AM-BN*) X CD. 

But, AM* - BN 2 is equal to (AM + BN) (AM - BN), 
(B. IV., P. X.) ; and because AM + BN is equal to 2 IK 
(P. IV., S.), and AM - BN to A 0, we have, 

vol CAB = \«IK x AG x CD. 

But, the right-angled triangles -A OB and CDI are simi- 
lar (B. IV., P. XXI.) ; hence, 

AO : AB : : CI : CD; or, AO x CD = AB x CI. 

Substituting, and changing the order of the factors, we have, 
vol. CAB = AB x 2*J0r x \CL 

But, AB x 2*IK is equal to the surface generated by 

A B ; hence, 

vol CAB = surf. AB x I CI. 

This demonstration holds good when the axis CD coin- 
cides with one side of the triangle CAB. 

2°. Suppose the base of the triangle to he parallel to 
the axis. 



BOOK VIII. 



229 




Draw AM and BjF perpendicular to the axis. The 
volume generated by CAB, is equal A T B 

to the cylinder generated by the rectan- 
gle ABJSfM, diminished by the sum of 
the cones generated by the triangles 
CAM and BCN '; hence, 

vol CAB = irCI 2 x AB - \«CI 2 x AI - \«C? x IB. 

But the sum of AI and IB is equal to AB : hence, 
we have, by reducing, and changing the order of the factors, 

vol. CAB = AB x 2«CI x I CI. 

But AB x 2* CI is equal to the surface generated by AB % 

consequently, 

vol. CAB - surf.AB x I CI; 

hence, in all cases, the volume generated by CAB is equal 
to surf. AB x i CI ; which was to be proved. 



PROPOSITION XIII. LEMMA. 

If a regular semi-polygon he revolved about its axis, the 
volume generated will he equal to the surface generated 
by the semi-perimeter multiplied by one-third of the 
apothem. 

Let FBDG be a regular scmi-poly- 
igon, FG its axis, 01 its apothem, and 

let the semi-polygon be revolved about 

FG : then will the volume generated 

be equal to surf.FBBG X {01 

For, draw lines from the vertices to 

the centre 0. These lines will divide 

the semi-polygon into isosceles triangles 

whose bases are sides of the semi-polygon, 

and whose altitudes are equal to 01 




230 



GEOMETRY. 



Now, the sum of the volumes generated by these trian- 
gles is equal to the volume generated by the semi-polygon. 
But, the volume generated by any triangle, as OAB, is 
equal to surf. AB x \OI (P. XH.) : hence, the volume 
generated by the semi-polygon is equal to surf. FBDG X \OI'> 
which was to be proved. 

Cor. The volume generated by a portion of the semi- 
polygon, OAJBC, limited by radii OC, OA, is equal to 
surf. ABC X 1 01. 



PEOPOSITIOK XIV. THEOBEM. 

The volume of a sphere is equal to its surface multiplied 

by one-third of its radius. 

Let ACE be a semicircle, AE its 
diameter, its centre, and let the semi- 
circle be revolved about AE: then will 
the volume generated be equal to the 
surface generated by the semi-circumfer- 
ence multiplied by one-third of the radius 
OA. 

For, the semicircle may be regarded 
as a regular semi-polygon having an infi- 
nite number of sides, whose semi-perimeter 

coincides with the semi-circumference, and whose apothem is 
equal to the radius : hence (P. XIII.), the volume gene- 
rated by the semicircle is equal to the surface generated by 
the semi-circumference multiplied by one-third of the radius ; 
which was to be proved. 

Cor. 1. Any portion of the semicircle, as OBC, bounded 
by two radii, will generate a volume equal to the surface 







BOOK VIII. 231 

generated by the arc BG multiplied by one-third of the 
radius (P. XIII,, C). But this portion of the semicircle is 
a circular sector, the volume which it generates is a spheri- 
cal sector, and the surface generated by the arc is a zone : 
hence, the volume of a spherical sector is equal to the zone 
which forms its base multiplied by one-third of the radius 

Cor. 2. If we denote the volume of a sphere by V, 
and its radius by E, the area of the surface will be equal 
to ±« R 2 (P. X., C. 1), and the volume of the sphere will be 
equal to 4*i2 2 X £i2; consequently, we have, 

V= i*B 3 . 

Again, it' we denote the diameter of the sphere by J9, we 
shall have B equal to |_Z>, and B? equal to }B\ and 
consequently, 

hence, the volumes of spheres are to each other as the cubes 
of their radii, or as the cubes of their diameters. 

Scholium. If the figure EBDE, 
formed bv drawing linos from the ex- 
tremities of the arc BD perpendicular 
to CA, be revolved about GA, as an 
axis, it will generate a segment of a 
sphere whose volume may be found by 
adding to the spherical sector generated 

by CDB, the cone generated by CBE, and subtracting 
from their sum the cone generated by CDE. If the arc 
BD is so taken that the points E and F fall on oppo- 
site sides of the centre (7, the latter cone must be added, 
instead of subtracted : hence, 




segment EBDE= zone BDx\ CD+^BE 2 x \ t 'E-*DF~ x \ CE 



232 



GEOMETRY. 



PROPOSITION" XV. THEOREM 

The surface of a sphere is to the entire surface of t/ie 
circumscribed cylinder, including its bases, as 2 is to 3 ; 
and the volumes are to each other in the same ratio. 

• 

Let PMQ be a semicircle, and PADQ a rectangle, 
whose sides PA and QD are tangent to the semicircle at 
P and Q, and whose side AD, is tangent to the semi- 
circle at If. If the semicircle and the rectangle "be revolved 
about PQ, as an axis, the former will generate a sphere, 
and the latter a circumscribed cylinder. 

1°. The surface of the sphere is to the entire surface oi 
the cylinder, as 2 is to 3. 

For, the surface of the sphere is 
equal to four great circles (P. X., C. 1), 
the convex surface of the cylinder is 
equal to the circumference of its base 
multiplied by its altitude (P.- I.) ; 
that is, it is equal to the circumfer- 
ence of a great circle multiplied by 
its diameter, or to four great circles 
"(B. V., P. XV.) ; adding to this the 

two bases, each of which is equal to a great circle, we have 
the entire surface of the cylinder equal to six great circles : 
hence, the surface of the sphere is to the entire surface of 
he circumscribed cylinder, as 4 is to C, or as 2 is to 3 ; 
which was to be proved. 




no 

- < 



The volume of the sphere is to the volume of the 
cylinder as 2 is to 3. 

For, the volume of the sphere is equal to %*P 3 (P. XIV., 
C. 2) ; the volume of the cylinder is equal to its base 
multiplied by its altitude (P. II.) ; that is, it is equal to 



BOOK VIII. 233 

*R 2 x 2i?, or to f fit 3 : hence, the volume of the sphere 
is to that of the cylinder as 4 is to 6, or as 2 is to 3 ; 
which teas to he proved. 

Cor. The surface of a sphere is to the entire surface of 
a circumscribed cylinder, as the volume of the sphere is to 
volume of the cylinder. 

Scholium. Any polyedron which is circumscribed about a 
sphere, that is, whose faces are all tangent to the sphere, 
may be regarded as made up of pyramids, whose bases are 
the faces of the polyedron, whose common vertex is at the 
centre of the sphere, and each of whose altitudes is equal 
to the -radius of the sphere. But, the volume of any one 
of these pyramids' is equal to its base multiplied by one- 
third of its altitude : hence, the volume of a circumscribed 
polyedron is equal to its surface multiplied by one-third of 
the radius of the inscribed sphere. 

Xow, because the volume of the sphere is also equal to 
its surface multiplied by one-third of its radius, it follows 
that the volume of a sphere is to the volume of any cir- 
cumscribed polyedron, as the surface of the sphere is to the 
surface of the polyedron. 

Polyedrons circumscribed about the same, or about equal 
spheres, are proportional to their surfaces. 



GEKB 1 1 AL FORMULAS. 



If we denote the convex surface of a cylinder by 5, its 
volume by "T, the radius of its base by 7?, and its alti- 
tude by 17", we have (P. I., II.), 

8 == 2irB x S (1-) 

V = <^ 2 x II ( 2 -) 



234 GEOMETRY. 

If we denote the convex surface of a cone by S, its 
volume by V, the radius of its base by B, its altitude by H, 
and its slant height by H\ we have (P. III., V.), 

B = «Bxflr ......... • (3.) 

V= «B*X i& (4.) 

If we denote the convex surface of a frustum of a cone 
hj JS, its volume by T 7 ", the radius of its lower base by B, 
the radius of its upper base by B\ its altitude by H, and its 
slant height by H\ we have (P. IV., VI.), 

S = «(B+B') xH' • • (5.) 

V= i^B* + B r * -\- B x 22') x J2\ . . (6.) 

If we denote the surface of a sphere by S, its volume 
by V, its radius by B, and its diameter by D, we have 
(P. X., C. 1, XIV., C. 2, XIV., C. 1), 

8 = 4tfi2 2 (7.) 

V = %«B 3 = }«D ? ( 8 -) 

If we denote the radius of a sphere by i?, the area of 

any zone of the sphere by JS, its altitude by II, and the 

volume of the corresponding spherical sector by V, we 
shall have (P. X., C. 2), 

S ^ 2 «B X H . . . ( 9.) 

V = !*I2 2 x II (10.) 

If we denote the volume of the corresponding spherical 
segment by I 7 ", the radius of its lower base by B\ the 
radius of its upper base by i2", the distance of its lower 
base from the centre by II', and the distance of its upper 
base from the centre by 27", we have (P. XIV., S.), 

V= \«i?W x H+B"*x II'^B'x II') . . (11.) 



BOOK IX. 

SPHERICAL GEOMETRY. 

DEFINITIONS. 

1. A Spherical Axgle is an angle included between the 
arcs of two great circles of a sphere meeting at a point. 
The arcs are called sides of the angle, and the point at 
which they meet is called the vertex of the angle. 

The measure of a spherical angle is the same as that of 
the diedral angle included between the planes of its sides. 
Spherical angles may be acute, right, or obtuse. 

2. A Spherical Poljgon is a portion of the surface of 
a sphere bounded by arcs of three or more great circles. 
The bounding arcs are called sides of the polygon, and the 
points in which the sides meet are called vertices of the 
polygon. Each side is supposed to be less than a semi-cir- 

cumfercnce. 

Spherical polygons are classified in the same manner as 

plane polygons. 

3. A Spherical Triangle is a spherical polygon of three 

sides. 

Spherical triangles are classified in the same manner as 

plane triangles. 

4. A Luxe is a portion of the surface of a sphere 
bounded by semi-circnmferenccs of two great circles. 

5. A Spherical Wedge is a portion of a sphere bounded 
by a lune and two semicircles meeting in a diameter of the 
sphere. 






236 



GEOMETRY. 



6. A Sphekical Pyramid is a portion of a sphere 
bounded by a spherical polygon and sectors of circles whose 
common centre is the centre of the sphere. 

The spherical polygon is called the base of the pyramid, 
and the centre of the sphere is called the vertex of the 
pyramid. 

7. A Pole oe a Circle is a point on the surface of 
the sphere, equally distant from all the points of the cir 

cumference of the circle. 

I 

8. A Diagonal of a spherical polygon is an arc of a 
great circle joining the vertices of any two angles which are 
not consecutive. 



PEOPOSITIOH" I. THEOEEM. 



Any side of a spherical triangle is less than the sum of 

the other two. 

Let ABC be a spherical triangle situated on a sphere 
whose centre is : then will any side, as AB, be less 
than the sum of the sides AG and BC, 

For, draw the radii OA, OB, and 
C : these radii form the ectees of a 
triedral angle whose vertex is 0, and 
the plane angles included between them 
are measured by the arcs AB, AC, 
and BC (B. ILL, P. XVIL, Sch.). 
But any plane angle, as A OB, is less 
than the sum of the plane angles AOC 
and BOC (B. VL, P. XIX.): hence, 

the arc AB is less than the sum of the arcs AC and 
BC; which was to be proved. 








BOOK IX. 237 

Cor. 1. Any side AB, of a spherical polygon ABODE, 
is less than the sum of all the other sides. 

For, draw the diagonals AC and AB, dividing the 
polygon into triangles. The arc AB is less than the sum 
of AC and BC, the arc A C is 
less than the sum of AB and BC, 
and the arc AB is less than the 
sum of BE and EA ; hence, AB 
is less than the sum of BC, CB, 
BE, and EA. 

Cor. 2. The arc of a great circle joining any two points 

on the surface of a sphere, is less than the arc of a small 
circle joining the .same points. 

For, divide the arc of the small circle into equal parts, 
and through the extremities of each part pass the arc of a 
great circle. The arc of the great circle joining the given 
points will he less than the sum of these arcs (C. 1), what- 
ever may he their number. But when this number is infinite, 
the arcs of the great circle coincide with the corresponding 
arcs of the small circle, and their sum is equal to the entire 
arc of the small circle. 

Cor. 3. The shortest distance between two points on 
the surface of a sphere, is measured on the arc of a great 
circle joining them. 

PROPOSITION II. THEOREM. 

TJie sum of the aides of a spherical polygon is less than 

f 
the circumference of a great circle. 

Let ABCBE be a spherical polygon situated on a 
sphere whose centre is : then will the sum of its sides 
be less than the circumference of a great circle. 



238 



GEOMETRY. 



For, draw the radii OA, OB, OC, OB, and OE: 
these radii form the edges of a polyedral angle whose vertex 
is at 0, and the angles included between 
them are measured by the arcs AB, BC, 
CD, BE, and EA. But the sum of 
these angles is less than four right angles 
(B. VI., P. XX.) : hence, the sum of the 
arcs which measure them is less than the 
circumference of a great circle ; which was 
to be proved. 




PROPOSITION III. THEOREM. 

If a diameter of a sphere be drawn perpendicular to the 
plane of any circle of the sphere, its extremities will be 
poles of that circle. 

Let C be the centre of a sphere, FJSTG any circle of 
the sphere, and BE a diameter of the sphere, perpendicular 
to the plane of FNG : then will the extremities B and E, 
be poles of the circle FNG. 

The diameter BE, being 
perpendicular to the plane of 
FNG, must pass through 
the centre (B. VIXL, 
P. VH, C. 3). If arcs of 
great circles BJST, BE, BG, 
&c, be drawn from B to 
different points of the cir- 
cumference FNG, and chords 
of these arcs be drawn, these 
chords will be equal (B. VI., 

P. V.), consequently, the arcs themselves will be equal. But 
these arcs are the shortest lines that can be drawn from the 




BOOK IX. 239 

point D, to the different points of the circumference (P. I., 
C. 2) : hence, the point £>, is equally distant from all the 
points of the circumference, and consequently is a pole of 
the circle (D. 1). In like manner, it may be shown that 
the point E is also a pole of the circle : hence, both B, 
and E, are poles of the circle FXG- ; which teas to be 
provecT. 

Cor. 1. Let AMB he a great circle perpendicular to 
BE: then will the angles DOM, ECJf, &c, be right 
angles ; and consequently, the arcs D3T, EM, &c, will 
each be equal to a quadrant (B. HI., P. XYIL, S.) : hence, 
the two poles of a great circle are at equal distances from 
the circumference. 

Cor. 2. The two poles of a small circle are at unequal 
distances from the circumference, the sum of the distances 
being equal to a semi-circumference. 

Cor. 3. The line DC being perpendicular to the plane 
AMB, any plane, as BMC, passed through it, Trill also 
be perpendicular to the plane AMB : hence, the spherical 
angle DMA, is a right-angle ; that is, if any point, in the 
circumference of a great circle, be joined with either pole by 
the arc of a great circle, such arc will be perpendicular to 
the circumference of the given circle. 

Cor. 4. If the distance of a point B, from each of the 
points A and M, in the circumference of a great circle, 
is equal to a quadrant, the point B, is the pole of the 

arc A M. 

For, let C be the centre of the sphere, and draw the 
radii CB, CA, CM. Since the angles A CD, MCB, are 
right angles, the line CB is perpendicular to the two 
Straight lines CM, CM: it is, therefore, perpendicular to their 



24:0 



GEOMETRY. 



plane (B. VI., P. IV.) : hence, the point B, is the pole of 
the arc AM. 

Scholium. The properties of these poles enable us to 
describe arcs of a circle on the surface of a sphere, with 
the same facility as on a plane surface. For, by turning 
the arc BE about the point B, the extremity F will 
describe the small circle FJS r G ; and by turning the quad- 
i ant JDFA round the point B, its extremity A will 
describe an arc of a great circle. 



PROPOSITION IY. THEOREM. 

The angle formed by two arcs of great circles, is equal to 
that formed by the tangents to these eves at their point 
of intersection, and is measured by the arc of a great 
circle described from the vertex as a pole, and limited 
by the sides, produced if necessary. 

Let the angle BAG be formed by the two arcs AB, 
AC: then is it equal to the angle FAG formed by the 
tangents AF, AG, and is measured by the arc BE of 
a great circle, described about A as a pole. 

For, the tangent AF, drawn in the 
plane of the arc AB, is perpendicular 
to the radius A ; and the tangent 
AG, drawn in the plane of the arc 
AG, is perpendicular to the same radius 
A : hence, the angle FA G is equal 
to the angle contained by the planes 
ABBH, ACER (B. VI., D. 4) ; which 
is that of the arcs AB, AG. Now, if 
the arcs AB and AF are both quad- 
rants, the lines GB, GE, are perpendicular to OA, and 




BOOK IX. 



241 



the angle DOE is equal to the angle of the planes ABDH, 
AC EH: hence, the arc DE is the measure of the angle 
contained by these planes, or of the angle CAB ; ichich 
toas to be proved. 

Cor. 1. The angles of spherical triangles may be com* 
pared by means of the arcs of great circles described from 
their vertices as poles, and included between their sides. 

A. spherical angle can always be constructed equal to a 
given spherical angle. 

Cor. 2. Vertical angles, such as 
ACO and BCN are equal; for 
either of them is the angle formed 
by the two planes ACE, OCJST. 
When two arcs ACE, OCJST, in- 
tersect, the sum of two adjacent 
angles, as ACO, OCB, is equal 
to two right angles. 




PROPOSITION V. THEOREM. 



If from tne vertices of the angles of a spherical triangle^ 
as poles, arcs be described forming a spherical triangle^ 
the vertices of the angles of this second triangle will be 
respectively poles of the sides of the first. 

From the vertices A, E, C, 
as poles, let the arcs EF, FE, 
ED, be described, forming the 
triangle DFE : then will the 
vertices D, E, and F, be 
respectively poles of the sides 
BC, AC, AE. 

For, the point A being 

16 




242 



GEOMETRY. 



the pole of the arc EF, the distance AE, is a quadrant ; 
the point G being the pole of the arc BE, the distance 
GE, is likewise a quadrant : hence, the point E is at a 
quadrant's distance from the points A and G : hence, it is 
the pole of the arc AG (P. HI., C. 4). It may he shown, 
in like manner, that D is the pole of the arc EG, and 
F that of the arc AB ; which teas to be proved. 

Scholium. The triangle ABC, may be described by 
means of BEE, as BEE is described by means of ABG. 
Triangles thus related are called polar triangles, or supple- 
mental triangles. 



PROPOSITION VI. THEOREM. 

Any angle, in one of two polar triangles, is measured by a 
semi-circumference, minus the side lying opposite to it in 
the other triangle. 

Let AB C, and EFB, be any two polar triangles : 
then will any angle in either triangle be measured by a 
semi-circumference, minus the side lying opposite to it in the 
other triangle. 

For, produce the sides AB, 
AG, if necessary, till they 
meet EF, in G and H. The 
point A being the pole of 
the arc GH, the angle A is 
measured by that arc (P. IV.). 
But, since E is the pole of 
All, the arc EH is a quad- 
rant ; and since F is the 

pole of AG, EG is a quadrant: hence, the sum of the 
;arcs EII and GF, is equal to a semi-circumference. But, 




BOOK IX. 243 

tiie sum of the arcs EH and GF, is equal to the sum 
of the arcs EF and Gil : hence, the arc GH, which 
measures the angle A, is equal to a semi-circumference, 
raiaus the arc EF. In like manner, it may be shown, that 
any other angle, in either triangle, is measured by a semi- 
circumference, minus the side lying opposite to it in the 
other triangle ; which was to be proved. 

Scholium. Besides the triangle JDEF, /\ 

three others may be formed by the / \ 

intersection of the arcs BE, EF, DF. x*^f~^^\/ 

But the proposition is applicable only -/ \ y\ 

to the central triangle, which is dis- % ^^«—\r \ 

tinguished from the other three by the v, -»— _jfo 

circumstance, that the two vertices, A 

and B, lie on the same side of B C ; the two vertices, 
B and E, on the same side of A C ; and the two verti- 
ces, C and F, on the same side of AJB. 



PROPOSITION VII. THEOREM. 

If from the vertices of any two angles of a spherical tri- 
angle, as poles, arcs of circles be described passing 
through the vertex of the third angle / and if from the 
second point in which these arcs intersect, arcs of great 
circles be drawn to the vertices, used as poles, the parts 
of the triangle thus formed will be equal to those of the 
given triangle, each to each. 

Let ABC be a spherical triangle situated on a sphere 
whose centre is 0, CED and CFD arcs of circles 
described about B and A as poles, and let DA and 
DB be arcs of great circles : then will the parts of the 



244: 



GEOMETRY. 




triangle ABB be equal to those of the given triangle 

ABC, each to each. 

For, by construction, the side AD 

is equal to AG, the side DB is 

equal to BG, and the side AB is 

common : hence, the sides are equal, 

each to each. Draw the radii OA, 
OB, OC, and OB. The radii OA, 
OB, and OC, will form the edges 

of a triedral angle whose vertex is 
; and the radii OA, OB, and OB, will form the 

edges of a second triedral angle whose vertex is also at ; 
and the plane angles formed by these edges will be equal, 
each to each : hence, the planes of the equal angles are 
equally inclined to each other (B. VI., P. XXI.). But, the 
angles made by these planes are equal to the corresponding 
spherical angles; consequently, the angle BAB is equal to 
BAG, the angle ABB to ABO, and the angle ABB 
to ACB: hence, the parts of the triangle ABB are equal 
to the parts of the triangle A GB, each to each; which 
was to be proved. 

Scholium 1. The triangles ABC and ABB, are not, 
in general, capable of superposition, but their parts are 
symmetrically disposed with respect to AB. Triangles 
which can be so placed are called symmetrical triangles. 

Scholium 2. If symmetrical triangles are isosceles, they 
can be so placed as to coincide throughout: hence, they 
are equal in area. 



BOOK IX. 



245 



PROPOSITION" VIII. THEOREM. 

If two spherical triangles, on the same, or on equal spheres, 
have two sides and the included angle of the one equal 
to two sides and the included angle .of the other, each 
to each, the remaining parts are equal, each to each. 

Let the spherical triangles ABC and EFG, have the 
side EF equal to AB, the side EG equal to AC, and 
the angle FEG equal to BAC : then will the side EG be 
equal to BC, the angle EFG to ABC, and the angle 
EGF to ACB. 

For, the triangle EFG may 
be placed upon ABC, or upon 
its symmetrical triangle AJDB, so 
as to coincide with it throughout, 
as may be shown by the same 
course of reasoning as that em- 
ployed in Book I., Proposition V. : 
hence, the side EG is equal to 

BC, the angle EFG to ABC, and the angle EGF to 
ACB; ichich icas to he proved. 




C G 




PROPOSITION IX. THEOREM. 

9 If two spherical triangles on the same, or on equal spheres, 
have two angles and the included side of the one equal 
to two angles and the included side of the other, each 
to each, the remaining parts will be equal, each to each 

Let the spherical triangles ABC and EFG, have the 
angle FEG equal to BAC, the angle EFG equal to 
ABC, and the side EF equal to AB : then will the 



246 



GEOMETRY. 



side EG be equal to AC, the side EG to BC, and 
the angle FGE to BCA. 

For, the triangle EFG may 
be placed npon ABC, or upon 
its symmetrical triangle ABB, so 
as to coincide with it throughout, 
as may be shown by the same 
course of reasoning as that em- 
ployed in Book I., Proposition 
VI. : hence, the side EG is equal 
to AC, the side EG to BC, 
BCA; which was to be proved. 




C G 




and the angle FGE to 



PROPOSITION X. THEOREM. 

If two spherical triangles on the same, or on equal spheres, 
have their sides equal, each to each, their angles will be 
equal, each to each, the equal angles lying opposite the 
equal sides. 

Let the spherical triangles EFG and ABC have the 
side EF equal to AB, the side EG equal to AC, and 
the side EG equal to BC: then will the angle FEG be 
equal to BAC, the angle EFG to ABC, and the angle 
EGF to ACB, and the equal angles will lie opposite the 
equal sides. 

For, it may be shown by the 
same course of reasoning as that 
employed in B. I., P. X., that the 

triangle EFG is equal in all 

° • D\ / ^C G 

respects, either to the triangle 

ABC, or to its symmetrical tri- 
angle ABB : hence, the angle 
FEG, opposite to the side EG, is equal to the angle BAG, 





BOOK IX. .247 

opposite to BC; the angle EFG, opposite to EG, is equal 
to the angle ABC, opposite to AG ; and the angle EGF, 
opposite to EF, is equal to the angle A CB, opposite to 
AB ; which was to be proved. 



PROPOSITION XI. THEOREM. 

i 

In any isosceles spherical triangle, the angles opposite the 
equal sides are equal ; and conversely, if two angles of 
a spherical triangle are equal, the triangle is isosceles. 

1°. Let ABG be a spherical triangle, having the side 
AB equal to AG: then will the angle G be equal to 
the angle B. 

For, draw the arc of a great circle 
from the vertex A, to the middle point 
D, of the base BG : then in the two 
triangles ABB and ABG, we shall have 
the side AB equal to AG, by hypothe- 
sis, the side BB equal to BG, by con- 
struction, and the side AB common ; 
consequently, the triangles have their angles equal, each to 
each (P. X.) : hence, the angle G is equal to the angle 
B ; which toas to be proved. 




2°. Let ABG be a spherical triangle having the angle 
G equal to the angle B : then will the side AB be 
equal to the side AG, and consequently the triangle wil 

be isosceles. 

For, suppose that AB and AC are not equal, but that 
one of them, as AB, is the greater. On AB lay off the 
arc BO equal to AG, and draw the arc of a great circle 
from to C : thou in the triangles ACB and OBC, 
we shall have the side AC equal to OB, by construction, 



248 



GEOMETRY. 




the side BO common, and the included angle A OB equal 
to the included angle OBC, by hypothesis : hence, the 
remaining parts of the triangles are equal, 
each to each, and consequently, the angle 
OCB is equal to the angle ABC. But, 
the angle A CB is equal to ABC, by 
hypothesis, and therefore, the angle OCB 
is equal to A OB, or a part is equal to 
the whole, which is impossible : hence, the 
supposition that AB and AC are un- 
equal, is absurd; they are therefore equal, and consequently, 
the triangle AB C is isosceles ; which was to be proved. 

Cor. The triangles ABB and ABC, having all of 
their parts equal, each to each, the angle ABB is equal 
to AB C, and the angle BAB is equal to BA C ; that 
is, if an arc of a great circle he drawn from the vertex 
of an isosceles spherical triangle to the middle of its base, 
it will be perpendicular to the base, and will bisect the verti- 
cal angle of the triangle. 



PROPOSITION XII. THEOREM. 

In any spherical triangle, the greater side is opposite the 
greater angle ; and conversely, the greater angle is oppo- 
site the greater side. 

1°. Let ABC be a spherical triangle, in which the angle 
A is greater than the angle B : then will the side B 
be greater than the side A C. 
For, draw the arc AB, 
making the angle BAB equal 
to ABB : then will AB be 
equal to BB (P. XL). But, 
the sum of AB and BO is 




BOOK IX t 



219 



greater than AC (P. I.); or, putting for AD its equal 
DD, we hare the sum of DD and DC, or DC, greater 
than A C ; which icas to be proved. 

2°. In the triangle ADC, let the side DC be greater 
than AC : then will the angle A be greater than the 
angle D. 

For, if the angles A and D were equal, the sides DC 
and AC would be equal ; or if the angle A was less 
than the angle D, the side DC would be less than AC, 
either of which conclusions is contrary to the hypothesis : hence, the 
angle A is greater than the angle B ; which was to be 'proved. 



PROPOSITION XIII. TIIEOEEM. 

If two triangles on the same, or on equal spheres, are 
mutually equiangular, they are also mutually equilateral. 

Let the spherical triangles A and D, be mutually equi- 
angular : then will they also be mutually equilateral. 

For, let P be the polar triangle of A, 
and Q the polar triangle of D : then, be- 
cause the triangles A and D are mutually 
equiangular, their polar triangles D and Q, 
must be mutually equilateral (P. VI.), and con- 
sequently mutually equiangular (P. X.). But, 
the triangles P and Q being mutually equi- 
angular, their polar triangles A and D, are 
mutually equilateral (P. VI.) ; which was to be proved. 

Scholium. This proposition does not hold good for piano 
triangles, for all similar plane triangles arc mutually equi- 
angular, but not necessarily mutually equilateral. Two 
spherical triangles on the same or on equal spheres, cannot 

i 

be similar without being equal. 




250 GEOMETRY. 



PEOPOSITION XIV. THEOEEM. 

The sum of the angles of a spherical triangle is less than 
six right angles, and greater than two right angles. 

Let ABC be a spherical triangle, and BEF its polar 
triangle : then will the sum of the angles A, B, and (7, 
be less than six right angles and greater than two. 

For, any angle, as A, be- 
ing measured by a semi-cir- 
cumference, minus the side 
EF (P. VI.), is less than two 
right angles: hence, the sum 
of the three angles is less than 
six right angles ; and because 
the measure of each angle is 
equal to a semi-circumference, 
minus the side lying opposite 

to it, in the polar triangle, the measure of the sum of the 
three angles is equal to three semi-circumferences, minus the 
sum of the sides of the polar triangle BEF But the 
latter sum is less than a circumference ; consequently, the 
measure of the sum of the angles A, B, and (7, is 
greater than a semi-circumference, and therefore the sum of 
the angles is greater than two right angles : hence, the sum 
of the angles A i B, and (7, is less than six right angles, 
and greater than two ; ichich was to be proved. 

Cor. 1. The sum of the three angles of a spherical tri- 
angle is not constant, like that of the angles of a rectilineal 
triangle, but varies between two right angles and six, with- 
out ever reaching either of these limits. Two angles, there- 
fore, do not serve to determine the third. 





.BOOK IX. 251 

Cor. 2. A spherical triangle may have two, or even three 
of its angles right angles : also two, or even three of its 
angles obtuse, 

.Cor. 3. If the triangle ABC is bi-rectan- 
gular, that is, has two right angles B and 
(7, the opposite sides of the polar triangle 
will be quadrants, and their point of intersec- 
tion will be the pole of the other side (P. 
III., C. 4). The angles opposite the equal 
sides are right angles (P. III., C. 3) : hence, the sides AB 
and A C are quadrants. 

If the angle A is also a right angle, the triangle ABC 
is tri-r octangular ; each of its angles is a right angle, and 
its sides are quadrants. Four tri-rectangular triangles make 
up the surface of a hemisphere, and eight the entire surface 

of a sphere. 

Scholium. The right angle is taken as the unit of mea- 
sure of spherical angles, and is denoted by 1. 

The excess of the sum of the angles of a spnerical tri- 
angle over two right angles, is called the spherical excess. 
II we denote the spherical excess by E, and the three 
angles expressed in terms of the right angle, as a unit, by 
A, B, and (7, we shall have, 

E=A + B+C-2. 

The spherical excess of any spherical polygon is equal to 
the excess of the sum of its angles over two right angles 
taken as many times as the polygon has sides, less two. 
If we denote the spherical excess by E> the sum of the 
angles by £, and the number of sides by w, we shall 
have, 

E = S — 2(n - 2) = S - 2n + 4. 



252 



GEOMETRY. 



PROPOSITION XV. THEOREM. 

Any lime, is to the surface of the sphere, as the angle 

of the lune is to four right angles, or as the arc which 

measures that angle is to the circumference of a great 
circle. 

Let A3IPJST be a lime, and MGJST the angle of the lune 
then will the area of the lune be to the surface of the sphere, 
as the arc 3IJST is to the circumference of a great circle 
MNjPQ; or, which is the same thing, as the angle 31 GJSf is 
to four right angles. 

In the first place, suppose the arc 
31JV and the circumference 3INPQ 
to be commensurable. For example, 
let them be to each other as 5 is 
to 48. Divide the circumference 
31JSFPQ into 48 equal parts, be- 
ginning at 31 ; 3IJST will contain 
five of these parts. Join each point 

of division with the points A and JB, by a quadrant : 
there will be formed 96 equal isosceles spherical triangles 
(P. VII., S. 2) on the surface of the sphere, of which the 
lune will contain 10 ': hence, in this case, the area of the 
lune is to the surface of the sphere, as 10 is to 96, or 
as 5 is to 48 ; that is, as the arc 3IJST is to the circum- 
ference 3IN~P Q, or as the angle of the lune is to four 
right angles. 

In like manner, the same relation may be shown to 
exist when the arc JLfiV, and the circumference 3IRPQ 
are to each other as any other whole numbers. 

If the arc MJST, and the circumference 3INPQ, are not 
commensurable, the same relation may be shown to exist by 




BOOK IX. 25 



o 



a course of reasoning entirely analogous to that employed in 
Book IV., Proposition III. Hence, in all cases, the area of 
a lune is to the surface of the sphere, as the angle of the 
lune is to four right angles, or as the arc which measures 
that angle is to the circumference of a great circle ; which 
was to be proved. 

Cor. 1. Lunes, on the same or on equal spheres, are to 
each other as their angles. 

Cor. 2. If we denote the area of a tri-rectangular triangle 

by T, the area of a lune by X, and the angle of the 

lune by A, the right angle being denoted by 1, we shall 

have, 

L : 8T : : A : 4; 
whence, 

L = T x 2A -, 

hence, the area of a lune is equal to the area of a tri- 
rectangular triangle multiplied by twice the angle of the 
lune. 

Scholium. The spherical wedge, whose angle is MCN^ 
is to the entire sphere, as the angle of the wedge is to four 
right angles, as may be shown by a course of reasoning 
entirely analogous to that just employed : hence, we infei 
that the volume of a spherical wedge is equal to the lune 
which forms its base, multiplied by one-third of the radius. 



PROPOSITION XVI. THEOREM. 
Symmetrical triangles are equal in area. 

Let ABC and BEE be symmetrical triangles, the 
side BE being equal to AB, the side BE to A C, and 
the side EE to BC : then will the triangles be equal in 
area. 



254 



GEOMETRY. 




For, conceive a small circle to be drawn through A, JB y 
and G, and let P be its pole ; draw arcs of great circles 
from P to A, B, and G: these 
arcs will be equal (D. 7). Draw 
the arc of a great circle FQ, 
making the angle DFQ equal to 
AGP, and lay off on it, FQ 
equal to GP\ draw arcs of great 
circles QD and QF. 

In the triangles PAG and 
FDQ, we have the side FD 

equal to AG, by hypothesis; the side FQ equal to PG, 
by construction, and the angle DFQ equal to AGP, by 
construction : hence (P. VIII.), the side DQ is equal to 
AP, the angle FDQ to PAG, and the angle FQD to 
J.PC. Now, because the triangles QFD and PJ.<7 are 
isosceles and equal in all their parts, they may be placed so 
as to coincide throughout, the side DF falling on AG, 
and the side QD on PA : hence, they are equal in area. 

If we take from the angle DFE the angle DFQ, and 
from the angle AGP the angle AGP, the remaining 
angles QFE and PGP, will be equal. In the triangles 
FQF and PGP, we have the side QF equal to PG, 
by construction, the side FE equal to PC, by hypothesis, 
and the angle QFE equal to PGP, from what has just 
been shown ; hence, the triangles are equal in all their 
parts, and being isosceles, they may be placed so as to 
coincide throughout, the side QE falling on PB, and the 
side QF on PG ; these triangles are, therefore, equal in 

area. 

In the triangles QDE and PAP, we have the sides 
QD, QE, PA, and PB, all equal, and the angle DQE 
equal to APB, because they are the sums of equal angles: 
hence, the triangles are equal in all their parts, and 



BOOK IX. 



255 



because they are isosceles, they may be so placed as to 
coincide throughout, the side QD foiling on PB, and the 
side QE on PA ; these triangles are, therefore, equal in 
area. 

Hence, the sum of the triangles QFP and QEE, is 
equal to the sum of the triangles PAG and PBC. If 
from the former sum we take away the triangle QPE, 
there will remain the triangle PEE; and if from the latter 
sum we take away the triangle PAP, there will remain 
the triangle ABC : hence, the triangles ABO and PEP 
are equal in area ; which was to be proved. 

Scholium. If the point P falls within the triangle ABC, 
the point Q will fall within the triangle PEE. In this 
case, the triangle PEE is equal to the sum of the triangles 
QFP, QFE, and QBE, and the triangle ABC is equal 
to the sum of the equal triangles PA G, PB G, and PAB ; 
the proposition, therefore, still holds good. 



PROPOSITION XVII. THEOREM. 

If the circumferences of two great circles int-ersect on the 
surface of a hemisphere, the sum of the opposite triangles 
thus formed, is equal to a lime whose angle is equal to 
that formed by the circles. 

Let the circumferences A OB, OOP, 
intersect on the surface of a hemis- 
phere : then will the sum of the oppo- 
site triangles A 00, B OP, be equal 
to the lune whose ani^le is BOP. 

For, produce the arcs OB, OP, 
on the other hemisphere, till they meet 
at JST. Now, since A OB and OBJST 
are semi-circumferences, if we take away the common part 




256 



GEOMETRY. 



OB, we shall have BN equal to AO. For a like roa- 
son, we have BN equal to CO, and BB equal to AC: 
hence, the two triangles AOC, BBJST, 
have their sides respectively equal : 
they are therefore symmetrical ; con- 
sequently, they are equal in area 
(P. XVI.). But the sum of the tri- 
angles BBJSf, BOB, is equal to 
the lune OBNBO, whose angle is 
B OB : hence, the sum of A C and 
BOB is equal to the lune whose 
angle is B OB ; which was to be proved. 

Scholium. It is evident that the two spherical pyramids, 
which have the triangles AOC, BOB. for bases, are 
together equal to the spherical wedge whose angle is BOB, 




PROPOSITION XVIII. THEOREM. 

The area of a spherical triangle is equal to its spherical 
excess multiplied by a tri-rectangular triangle. 

Let AB C be a spherical triangle : then will its surface 

be equal to 

(A + B + C - 2) x T. 

For, produce its sides till they meet 
the great circle BEFG, drawn at plea- 
sure, without the triangle. By the last 
theorem, the two triangles ABE, A GIT, 
are together equal to the lune whose 
anp-le is A ; but the area of this lune 
is equal to 2A X T (P. XV., C. 2) : 

hence, the sum of the triangles ABE and AGE, is equal 
to 2A X T. In like manner, it may be shown that the 




BOOK IX. 257 

sum of the triangles BFG and BID, is equal to 2B x T, 
and that the sum of the triangles CUT and GFJE, is 

equal to 2 C X T. 

But the sum of these six triangles exceeds the hemis- 
phere, or four times T, by twice the triangle ABC. We 
shall therefore have, 

2 x area ABC = 2A X T+ 2jB x ^+ 2(7 X T - AT ; 
or, by reducing and factoring, 

area ABC = (A + B + C - 2) X T ; 
which was to be proved. 

Scholium 1. The same relation which exists between the 
spherical triangle ABC, and the tri-rectangular triangle, 
exists also between the spherical pyramid which has ABC 
for its base, and the tri-rectangular pyramid. The triedral 
angle of the pyramid is to the triedral angle of the tri- 
rectangular pyramid, as the triangle ABC to the tri-rectan- 
gular triangle. From these relations, the following conse- 
quences are deduced : 

1°. Triangular spherical pyramids are to each other as 
their bases ; and since a polygonal pyramid may always be 
divided into triangular pyramids, it follows that any two 
spherical pyramids are to each other as their bases. 

2°. Polyedral angles at the centre of the same, or of 
equal spheres, are to each other as the spherical polygons 
intercepted by their faces. 

Scholium 2. A polyedral angle whoso faces are perpen- 
dicular to each other, is called a right polyedral angle ; 
and if placed at the centre of a sphere, its faces will inter- 
cept a tri-rectangular triangle. The right polyedral angle is 

17 



258 GEOMETRY. 

taken as the unit of polyedral angles, and the tri-rectangular 
spherical triangle is taken as its measure. If the vertex of 
a polyedral angle be taken as the centre of a sphere, the 
portion of the surface intercepted by its faces will be the 
measure of the polyedral angle, a tri-rectangular triangle of 
the same sphere, being the unit. 

PROPOSITION XIX. THEOREM. 

The area of a spherical polygon is eqical to its spherical 
excess multiplied by the tri-rectangular triangle. 

Let ABODE be a spherical polygon, the sum of whose 
angles is S, and the number of whose sides is n : then 
will its area be equal to 

(fl - 2n + 4) x T. 

For, draw the diagonals AC, AD, 
dividing the polygon into spherical tri- 
angles : there will be n — 2 such tri- 
angles. Now, the area of each tri- 
angle is equal to its spherical excess 
into the tri-rectangular triangle : hence, 
the sum of the areas of all the triangles, or the area of the 
polygon, is equal to the sum of all the angles of the tri- 
angles, or the sum of the angles of the polygon diminished 
by 2{n — 2) mto the tri-rectangular triangle ; or, 

area AJBCDJS = [S - 2{n - 2)] x T ; 
whence, by reduction, 

area ABCDE = (S - 2n + 4) x T ; 
which icas to be proved. 




BOOK IX. 259 



GENERAL SCHOLIUM. 

From any point on a hemisphere, two arcs of great circl 
can always be drawn which shall be perpendicular to the ci. 
cumference of the base of the hemisphere, and they will in 
general be unequal. JSTow, it may be proved, by a course oi 
reasoning analogous to that employed in Book I., Proposition 
XV.: 

1°. That the shorter of the two arcs is the shortest arc 
that can be drawn from the given point to the circum- 
ference '. 

2°. That two oblique arcs drawn from the same point, to 
points of the circumference at equal distances from the foot 
of the perpendicular, are equal : 

3°. That of two oblique arcs, that is the longer which 
meets the circumference at the greater distance from the foot 
of the perpendicular. 

This property of the sphere is used in the discussion of 
triangles in spherical trigonometry. 



TRIGONOMETRY 



AND 



MENSURATION. 



INTRODUCTION TO TRIGONOMETRY. 



LOGARITHMS. 



1. The Logarithm of a number is the exponent of the 
power to which it is necessary to raise a fixed number, to 
produce the given number. 

The fixed number is called the base of the system. Any 
positive number, except 1, may be taken as the base of a 
system. In the common system, the base is 10. 

2. If we denote any positive number by n, and the 

corresponding exponent of 10, by cc, we shall have the 

exponential equation, 

10* = n (1.) 

In this equation, x is, by definition, the logarithm of n> 
which may be expressed thus, 

x = log n (2.) 

3. From the definition of a logarithm, it follows that, the 
logarithm of any power of 10 25 equal to the exponent of 
that power : hence the formula, 

log(10) p = p (3.) 

If a number is an exact power of 10, its logarithm is 
a whole number. 



4 INTRODUCTION. 

If a number is not an exact power of 10, its logarithm 
will not be a whole number, but will be made up of an 
entire part plus a fractional part, which is generally expres- 
sed decimally. The entire part of a logarithm is called the 
characteristic, the decimal part, is called the mantissa. 

4. If, in Equation ( 3 ), we make p successively equal 
to 0, 1, 2, 3, &c, and also equal to — 0, — 1, — 2, — 3, 
&c, we may form the following 









TABLE. 






log 1 


— 











log 10 


^^ 


1 


log .1 


— 


- 1 


log 100 


— 


2 


log .01 


— 


— 2 


log 1000 


— 


3 


log .001 


= 


- 3 


&c, 


&c. 




&c, 


&c. 





If a number lies between 1 and 10, its logarithm lies 
between and 1, that is, it is equal to plus a deci- 
mal ; if a number lies between 10 and 100, its logarithm 
is equal to 1 plus a decimal ; if between 100 and 1000, 
its logarithm is equal to 2 plies a decimal ; and so on : 
hence, we have the following 

EULE. 

The characteristic of the logarithm of an entire number is 
positive, and numerically 1 less than the number of places 
of figures in the given number. 

If a decimal fraction lies between .1 and 1, its loga- 
rithm lies between — 1 and 0, that is, it is equal to — 1 
plus a decimal ; if a number lies between .01 and .1, its 
logarithm is equal to — 2, plus a decimal ; if between .001 
and .01, its logarithm is equal to — 3, plus a decimal ; 
and so on : hence, the following 



TRIGONOMETRY. 5 



RULE. 



The characteristic of the logarithm of a decimal fraction 
is negative, and numerically 1 greater than the number 
of O's that immediately follow the decimal point. 

The characteristic alone is negative, the mantissa being 
always positive. This fact is indicated by writing the neg- 
ative sio-n over the characteristic : thus, 2.371465, is equiv- 
alent to — 2 + .371465. 

It is to be observed, that the characteristic cf the logarithm 
of a mixed number is the same as that of its entire part. 
Thus, the mixed number 74.103, lies between 10 and 100; 
hence, its logarithm lies between 1 and 2, as does the logarithm 
of 74. 



GENERAL PRINCIPLES. 
V 

5. Let m and n denote any two numbers, and x 
and y their logarithms. We shall have, from the defini- 
tion of a logarithm, the following equations, 

10* = m (4.) 

10 y = n (5.) 

Multiplying (4) and (5), member by member, we have, 

10 r + y = win; 

whence, by the definition, 

x + y = log (mn) (6.) 

That is, the logarithm of the product of two numbers is 
equal to the sum of the logarithms of tlie numbers. 



6 INTRODUCTION. 

6. Dividing ( 4 ) "by ( 5 ), member by member, we have, 

10*-' = m ; 
« n 

whence, by the definition, 

x-y = log(^J (7.) 

That is, the logarithm of a quotient is equal to the loga- 
rithm of the dividend diminished by that of the divisor. 

7. Raising both members of (4) to the power denoted 

by Pi we have, 

10^ = m p ; 

whence, by the definition, 

xp = log m v (8.) 

That is, the logarithm of any power of a number is equal 
to the logarithm of the number multiplied by the exponent 
of the power. 

8. Extracting the root, indicated by r, of both members 
of (4), we have, 

X 

lCT = \fm ; 
whence, by the definition, 

® = log'^/nT. .... (9.) 

That is, the logarithm of any root of a number is equal 
to the logarithm of the number divided by the index of the 
root. 

The preceding principles enable us to abbreviate the oper 
ations of multiplication and division, by converting them into 
'the simpler ones of addition and subtraction. 



TRIGONOMETRY. 



TABLE OF LOGARITHMS. 

9. A Table of Logakithms, is a table by means of 
which we can find the logarithm corresponding to any num- 
ber, or the number corresponding to any logarithm. 

In the table appended, the complete logarithm is given 
for all numbers from 1 up to 100. For other numbers, the 
mantissas alone are given; the characteristic may be found be 

one of the rules of- Art. 4. 

Before explaining the use of the table, it is to be shown 
that the mantissa of the logarithm of any number is not 
changed by multiplying or dividing the number by any exact 
power of 10. 

Let n represent any number whatever, and 10 ? any 
power of 10, p being any whole number, either positive 
or negative. Then, in accordance with the principles of Arts. 
5 and 3, we shall have, 

log {n x 10 p ) - log n + log 10 p = p + log n ; 

but p is, by hypothesis, a whole number : hence, the deci- 
mal part of the log (n x 10 p ) is the same as that of log n ; 
which teas to be proved. 

Hence, in finding the mantissa of the logarithm of a num- 
ber, w T e may regard the number as a decimal, and move the 
decimal point to the right or left, at pleasure. Thus, the 
mantissa of the logarithm of 45G357, is the same as that of 
the number 45G3.57 ; and the mantissa of the logarithm of 
2.0035 7, is the same as that of 2003.57. 



8 INTRODUCTION. 

) MANNER OE USING THE TABLE. 

i 

1°. To find the logarithm of a number less than 100. 

10. Look on the first page, in the column headed "N," 
fbr the given number ; the number opposite is the logarithm 
required. Thus, 

log 67 = 1.826075. 

2°. To find the logarithm of a number between 100 and 

10,000. 

11. Find the characteristic by the first rule of Art. 4. 
To find the mantissa, look in the column headed "1ST," 

for the first three figures of the number ; then pass along 
a horizontal fine until you come to the column headed with 
the fourth figure of the number ; at this place will be found 
four figures of the mantissa, to which, two other figures, 
taken from the column headed "0," are to be prefixed. If 
the figures found stand opposite a row of six figures, in the 
column headed "0," the first two of this row are the ones 
to be prefixed ; if not, ascend the column till a row of six 
figures is found ; the first two, of this row, are the ones to 
be prefixed. 

If, however, in passing back from the four figures, first 
found, any dots are passed, the two figures to be prefixed 
must be taken from the line immediately below. If the 
figures first found fall at a place where dots occur, the dots 
must be replaced by 0's, and the figures to be prefixed must 
be taken from the line below. Thus, 

Log 8979 = 3.953228 
Log 3098 - 3.491081 
Log 2188 = 3.340047 



\ 



TRIGONOMETRY. 9 

3°. To find the logarithm of a number greater than 10,000. 

12. Find the characteristic by the first rule of Art. 4. 

To find the mantissa, place a decimal point after the fourth 
figure (Art. 9), thus converting the number into a mixed 
number. Find the mantissa of the entire part, by the me- 
thod last given. Then take from the column headed "D," 
the corresponding tabular difference, and multiply this by the 
decimal part and add the product to the mantissa just found 
The result will be the required mantissa. 

It is to be observed that when the decimal part of the 
product just spoken of is equal to or exceeds .5, we add 
1 to the entire part, otherwise the decimal part is rejected 

EXAMPLE. 

1. To find the logarithm of 672887. 

The characteristic is 5. Placing a decimal point after the 
fourth figure, the number becomes 6728.87. The mantissa 
of the logarithm of 6728 is 827886, and the corresponding 
number in the column "D" is 65. Multiplying 65 by .87, 
we have 56.55 ; or, since the decimal part exceeds .5, 57. 
We add 57 to the mantissa already found, giving 827943, 
and we finally have, 

log 672887 = 5.827943. 

* 

The numbers in the column "D" are the differences be- 
tween the logarithms of two consecutive whole numbers, and 
are found by subtracting the number mder the heading "4' 
from that under the heading "5." 

In the example last given, the mantissa of the logarithm 
of 6728 is 827886, and that of 6729 is 827951, and 
their difference is 65 ; 87 hundredths of this difference is 



10 INTRODUCTION. 

57 : hence, the mantissa of the logarithm of 6728.87 is found 
by adding 57 to 827886. The principle employed is, that 
the differences of numbers are proportional to the difference^ 
of their logarithms, when these differences are small. 



4°. To find the logarithm of a decimal. 

13. Find the characteristic by the second rule of Art. 4. 
To find the mantissa, drop the decimal point, thus reduc- 
ing the decimal to a whole number. Find the mantissa of 
the logarithm of this number, and it will be the mantissa 
required. Thus, 

log .0327 = 2.514548 
log 378.024 = 2.577520 



5°. To find the number corresponding to a given logarithm. 

14. The rule is the reverse of those just given. Look 
in the table for the mantissa of the given logarithm. If it 
cannot be found, take out the next less mantissa, and also 
the corresponding number, which set aside. Find the differ- 
ence between the mantissa taken out and that of the given 
logarithm ; annex as many 0's as may be necessary, and 
divide this result by the corresponding number in the colu 
"D." Annex the quotient to the number set aside, and th 
point off, from the left hand, a number of places of figure 
equal to the characterististic plus 1 : the result will be the 
number required. If the characteristic is negative, the result 
will be a pure decimal, and the number of 0's which im- 
mediately follow the decimal point will be one less than the 
number of units in the characteristic. 



TRIGONOMETRY. 11 

EXAMPLES. 

1. Let it be required to find the number corresponding 
to the logarithm 5.233568. 

The next less mantissa in the table is 233504 ; the cor- 
responding number is 1712, and the tabular difference is 
253. 

OPERATION. 

Given mantissa, 233568 

Next less mantissa, • • • 233504 - • 1712 

253 ) 6400000 ( 25296 

.'. The required mumber is 171225.296. 

The number corresponding to the logarithm 2.233568 is 
.0171225. 

2. What is the number corresponding to the logarithm 
2.785407 ? Ans. .06101084. 

3. What is the number corresponding to the logarithm 
1.846741 ? Ans. .702653. 



MULTIPLICATION BY MEANS OF LOGARITHMS. 

15. From the principle proved in Art. 5, we deduce the 
following 

rule . 

Find the logarithms of the factors, and taJce their sum , 
then find the number corresponding to the resxdting logarithm^ 
and it will be the product required. 



12 INTRODUCTION. 

EXAMPLES. 

1. Multiply 23.14 by 5.062. 

OPERATION. 

log 23.14 • • • 1.364363 
log 5.062 • • • 0.704322 

2.068685 .'. 117.1347, product. 

2. Find the continued product of 3.902, 597.16, and 
0.0314728. 

OPERATION. 

log 3.902 • • • 0.591287 

log 597.16 • • ■ 2.776091 
log 0.0314728 • • • 2.497936 

1.865314 .'. 73.3354, product. 

Here, the 2 cancels the + 2, and the 1 carried from 
the decimal part is set down. 

3. Find the continued product of 3.586, 2.1046, 0.8372, 
and 0.0294. -^s. 0.1857615. 



DIVISION BY MEANS OF LOGAKITHMS. 

16. From the principle proved in Art. 6, we have the 
following 

RULE. 

Find the logarithms of the dividend and divisor, and 
subtract the latter from the former; then find the number 
corresponding to the resulting logarithm, and it will be the 
quotient required. 



TRIGOjS 1 OMETRT. 



13 



EXAMPLES, 



1. Divide 24163 by 4567. 



log 24163 
log 4567 



OPERATION. 

• 4.383151 
- 3.659631 

0.723520 



5.59078, quotient. 



2 Divide 0.7438 by 12.9476. 



log 0.7438 • 
log 12.9476 • 



OPERATION. 

• 1.871456 

• 1.112189 

2.759267 



0.057447, quotient. 



Here, 1 taken from 1, gives 2 for a result. The 
subtraction, as in this case, is always to be performed in the 
algebraic sense. 



3. Divide 37.149 by 523.76. 



Ans. 0.0709274. 



The operation of division, particularly when combined with 
that of multiplication, can often be simplified by using the 
principle of 



THE ARITHMETICAL COMPLEMENT. 

17. The Arithmetical Complement of a logarithm is the 
result obtained by subtracting it from 10. Thus, 8.130456 
is the arithmetical complement of 1.809544. The arithmetical 
complement of a logarithm may be written out hy commenc- 
ing at the left hand and subtracting each figure from 9, 

18 



14: 



INTRODUCTION. 



until the last significant figure is reached, which must he 
taken from 10. The arithmetical complement is denoted by 
the symbol (a. c). 

Let a and b represent any two logarithms whatever, 
and a — b their difference. Since we may add 10 to, 
and subtract it from, a — b, without altering its value, we 
have, 

a — b = a + (10 - b) - 10. . . . ( 10.) 

But, 10 — 5 is, by definition, the arithmetical complement 
of b : hence, Equation ( 10 ) shows that the difference be- 
tween two logarithms is equal to the first, plus the arith- 
metical complement of the second, minus 10. 

Hence, to divide one number by another by means of 
the arithmetical complement, we have the following 

RULE. 

Find the logarithm of the dividend, and the arithmetical 
complement of the logarithm of the divisor, add them toge- 
ther, and diminish the sum by 10 ; the number correspond 
ing to the resulting logarithm will be the quotient required, 

EXAMPLES. 

1. Divide 327.5 by 22.07. 



log 327.5 
(a. c.) log 22.07 



OPERATION. 

• 2.515211 

• 8.656198 



1.171409 .'. 14.839, quotient 



2. Divide 37 149 by 523.76. 



Ans. 0.0709273. 



TRIGONOMETRY. 15 

3. Multiply 358884 by 5672, and divide the piodoct 
by 89721. 

OPERATION. 

log 358884 • • • 5.554954 

log 5672 • • • 3.753736 

(a. c.) log 89721 • • • 5.047106 

4.355796 .'. 22688, result. 



4. Solve the proportion, 

3976 : 7952 : : 5903 : x. 



OPERATION. 



log 7952 • • • 3.900476 

losr 5903 • • • 3.771073 

(a. c.) log 3976 • • • 6.400554 



4.072103 ,\ x = 11808 



The operation of subtracting 10 is always performed 
mentally. 

EAISIXG TO POWEES BY MEAN9 OF LOGAEITIHrS. 

18. From the principle proved in Art. 7, we have th€ 
following 

RULE. 

Find the logarithm of the number, and multiply it by 
the exponent of the power ; then find the number correspond- 
ing to the resulting logarithm, and it will be the power 
required. 



16 INTRODUCTION. 

EXAMPLES. 

lc Find the 5th power of 9. 

OPERATION. 



log 9 . . . 0.954243 

5 



4.771215 .». 59049, power. 
2. Find the 7th power of 8. Ans. 2097152. 

EXTRACTING ROOTS BY MEANS OF LOGARITHMS. 

19. From the principle proved in Art. 8, we have the 
following 

RULE. 

Find the logarithm of the number, and divide it by the 
mdex of the root ; then find the number corresponding to 
the resulting logarithm, and it will be the root required. - 1 

EXAMPLES. 

1. Find the cube root of 4096. 

The logarithm of 4096 is 3.612360, and one-third of 
this is 1.204120. The corresponding number is 10, which 
is the root sought. 

When the characteristic is negative and not divisible by 
the index, add to it the smallest negative number that will 
maize it divisible, and then prefix the same number, with a 
plus sign, to the mantissa. 

2. Find the 4th root of .00000081. 
The logarithm of .00000031 is V.908485, which is equal 

to 8 + 1.908485, and one-fourth of this is 2.477121. 

The number corresponding to this logarithm is 03 : 
hence, .03 is the root required. 



PLANE TRIGONOMETRY. 



20 Plane Trigonometry is that branch of Mathematics 
which treats of the solution of plane triangles. 

In every plane triangle there are six parts : three sides 
and three angles. When three of these parts are given, one 
being a side, the remaining parts may be found by comput- 
ation. The operation of finding the unknown parts, is called 
the solution of the triangle. 




21. A plane angle is measured by the arc of a circle 
included between its sides, the centre of the circle being at 
the vertex, and its radius being equal to 1. 

Thu3, if the vertex A be taken 
as a centre, and the radius AB be 
equal to 1, the intercepted arc B C 
will measure the angle A (B. III., P. 

XVII., s.). 

* Let ABCB represent a circle whose radius is equal to 
I, and AC, BB, two diameters per- 
pendicular to each other. These dia- 
meters divide the circumference into 
four equal parts, called quadrants ; and 
because each of the angles at the cen- 
tre is a right angle, it follows that a 
right angle is measured by a quad- 




18 



PLANE TRIGONOMETRY. 



rant. An acute angle is measured by an arc less than a 
quadrant, and an obtuse angle, by an arc greater than a 
quadrant. 



22. In Geometry, the unit of angular measure is a right 
angle ; so in Trigonometry, the primary unit is a quadrant, 
which is the measure of a right angle. 

For convenience, the quadrant is divided into 90 equal 
parts, each of which is called a degree ; each degree into 
60 equal parts, called minutes; and each minute into 60 
equal parts, called seconds. Degrees, minutes, and. seconds, 
are denoted by the symbols °, ', ". Thus, the expression 
7° 22' 33", is read, 7 degrees, 22 minutes, and 33 seco?ids. 
Fractional parts of a second are expressed decimally. 

A quadrant contains 324,000 seconds, and an arc of 7° 
22' 33" contains 26553 seconds; hence, the angle measured 
by the latter arc, is the AWW& part of a right angle. 
In like manner, any angle may be expressed in terms of a 
right angle. 



23. The complement of an arc is the difference between 
that arc and 90°. The complement 
of an angle is the difference be- 
tween that angle and a right angle. 

Thus, EB is the complement of 
AE* and FB is the complement 
of AF. In like manner, FOB 
is the complement of A OF, and 
FOB is the complement of A OF. 

In a right-angled triangle, the 
acute angles are complements of each other. 




24. The supplement of an arc is the difference between 



PLANE TRIGONOMETRY. 



19 



that arc and 180°. The supplement of an angle is the dif- 
ference between that angle and two right angles. 

Thus, EG is the supplement of AE, and EG the 
supplement of AF. In like manner, EOG is the supple- 
ment of AOE, and FOG the supplement of A OF. 

In any plane triangle, either angle is the supplement of 
the sum of the other two. 



25. Instead of employing the arcs themselves, we usually 
employ certain functions of the arcs, as explained below. 
A function of a quantity is something which depends upon 
that quantity for its value. 

The following functions are the only ones needed for solv- 
ing triangles : 

26. The sine of an arc is the distance of one extremity 
of the arc from the diameter, through the other extremity. 

Thus, PM is the sine of 
AM, and F'M' is the sine of 

AM'. 

If AM is equal to M' C, 
AM and AM' will be supple- 
ments of each other ; and be- 
cause MM' is parallel to AG, 
PM will be equal to F'M' 
(B. I., P. XXIII.) : hence, the 
sine of an arc is equal to the 
nine of its supplement. 

27. The cosine of an arc is the sine of the complement 
of the arc. 

Thus, NM is the cosine of AM, and NM % is the 
cosine of AM'. These lines are respectively equal to OP 
and OP'. 




Mil 



20 



PLANE TRIGONOMETRY. 



It is evident, from the equal triangles of the figure, that 
the cosine of an arc is equal to the cosine of its supple- 
ment. 




28. The tangent of an arc is the perpendicular to the 
radius at one extremity of the arc, limited by the prolon- 
gation of the diameter through the other extremity 

Thus, AT is the tangent of 
the arc A3I, and AT'" is 
the tangent of the arc AM'. 

If AM is equal to 31' C, 
AM and AM' will be supple- 
ments of each other. But AM"' 
and AM' are also supplements 
of each other : hence, the arc 
AM is equal to the arc AM'", 
and the corresponding angles, 

A 031 and A 031'", are also equal. The right-angled tri- 
angles AOT and AOT", have a common base AO, and 
the angles at the base equal ; consequently, the remaining 
parts are respectively equal : hence, AT is equal to AT'". 
But AT is the tangent of A3I, and AT'" is the tangent 
of AM' : hence, the tangent of an arc is equal to the tan- 
gent of its supplement. 

It is to be observed that no account is taken of the alge- 
braic signs of the cosines and tangents, the numerical values 
alone being referred to. 

29. The cotangent of an arc is the tangent of its com- 
plement. 

Thus, BT' is the cotangent of the arc A3I, and BT" 
is the cotangent of the are A3f. 

The sine, cosine, tangent, and cotangent of an arc, a, 
are, for convenience, Written sin a, cos a, tan a, and cot a. 



PLANE TRIGONOMETRY. 



21 



These functions of an arc have been defined on the sup- 
position that the radius of the arc is equal to 1 ; in this 
case, they may also be considered as functions of the angle 
which the arc measures. 

Thus, PM, NM, AT, and BT\ are respectively the 
sine, cosine, tangent, and cotangent of the angle A 031, as 
well as of the arc AM. 

30. It is often convenient to use some other radius than 
1 ; in such case, the functions of the arc, to the radius 1, 
may be reduced to corresponding functions, to the radius R. 

Let A 031 represent any angle, 
A3! an arc described from as 
a centre with the radius 1, PM 
its sine ; A' 31' an arc described 
from as a centre, with any ra- 
radius P, and P'M' its sine. 
Then, because OP 31 and OP'3f 
are similar triangles, we shall have, 

031 : P3I : : 031' : P'M', or, 1 : PM : : X : P'M' ; 

whence, 

P3I = ^2£- , and, P'M' = PM x P ; 

and similarly for each of the other functions. 

That is, any function of an arc whose radius is 1, is 
equal to the corresponding function of an arc whose radius 
is P, divided by that radius. Also, any function of an 
arc whose radius is 22, is equal to the corresponding func- 
tion of an arc whose radius is 1, multiplied by the ra- 
dius H. 

By making these changes in any formula, the formula will 

be rendered homogeneous. 




Y A' 



22 PLANE TRIGONOMETRY 

TABLE OF "NATUEAL SINES. 

31. A Natural Suns, Cosine, Tangent, ok Cotangent, 
is the sine, cosine, tangent, or cotangent of an arc whose 
radius is 1. 

A Table of Natural Sines is a table by means of which 
the natural sine, cosine, tangent, or sotangent of any arc, 
may be found. 

Such a table might be used for all the purposes of tri- 
gonometrical computation, but it is found more convenient to 
employ a table of logarithmic sines, as explained in the next 
article. 

TABLE OF LOGARITHMIC SINES. 

32. A Logarithmic Sine, Cosine, Tangent, or Cotan- 
gent is the logarithm of the sine, cosine, tangent, or cotan- 
gent of an arc whose radius is 10,000,000,000. 

A Table* of Logarithmic Sines is a table from which the 
logarithmic sine, cosine, tangent, or cotangent of any arc may 
be found. 

The logarithm of the tabular radius is 10. 

Any logarithmic function of an arc may be found by mul- 
tiplying the corresponding natural function by 10,000,000,000 
(Art. 30), and then taking the logarithm of the result ; or 
more simply, by taking the logarithm of the corresponding 
natural function, and then adding 10 to the result (Art. 5). 

33. In the table appended, the logarithmic functions are 
given for every minute from 0° up to 90°. In addition, 
their rates of change for each second, are given in the 
column headed "D." 

The method of computing the numbers in the column 
headed "D," will be understood from a single example. The 



PLANE TRIGONOMETRY. 23 

logarithmic sines of 27° 34', and of 27° 35', are, respect- 
ively, 9.665375 and 9.665617. The difference between their 
mantissas is 242 ; this, divided by 60, the number of sec- 
onds in one minute, gives 4.03, which is the change in the 
mantissa for 1", between the limits 27° 34' and 27° 35'. 

For the sine and cosine, there are separate columns of 
differences, which are written to the right of the respective 
columns ; but for the tangent and cotangent, there is but a 
sinoie column of differences, which is written between them. 
The logarithm of the tangent increases, just as fast as that 
of the cotangent decreases, and the reverse, their sum being 
always equal to 20. The reason of this is, that the product 
of the tangent and cotangent is always equal to the square 
of the radius ; hence, the sum of their logarithms must 
always be equal to twice the logarithm of the radius, or 20. 

The angle obtained by taking the degrees from the top 

of the page, and the minutes from airy line on the left hand 

of the page, is the complement of that obtamed by taking 

■ 
the degrees from the bottom of the page, and the minutes 

from the same line on the right hand of the page. But, 

by definition, the cosine and the cotangent of an arc are, 

respectively, the sine and the tangent of the complement of 

that arc (Arts. 26 and 28) : hence, the columns designated 

sine and tariff, at the top of the page, are designated cosine 

and cotang at the bottom. 

USE OF TIIE TABLE. 

To find the logarithmic functions of an arc ichich is ex- 
pressed in degrees and minutes. 

34. If the arc is less than 45°, 100k for the degrees at 
the top of the page, and for the minutes in the left hand 
column ; then follow the corresponding horizontal line till you 



24= PLANE TRIGONOMETRY. 

come to the column designated at the top by sine, cosine, 
tang, or cotang, as the case may be ; the number there 
found is the logarithm required. Thus, 

log sin 19° 55' • • . 9.532312 
log tan 19° 55' • • . 9.559097 

If the angle is greater than 45°, look for the degrees at 
the bottom of the page, and for the minutes in the right 
hand column ; then follow the corresponding horizontal line 
backwards till you come to the column designated at the bot- 
tom by sine, cosine, tang, or cotang, as the case may be ; 
the number there found is the logarithm required. Thus, 

log cos 52° 18' • • • 9.786416 

log tan 52° 18' '• • . 10.111884 

To find the logarithmic functions of an arc which is ex- 
pressed in degrees, minutes, and seconds. 

35. Find the logarithm corresponding to the degrees and 
minutes as before ; then multiply the corresponding number 
taken from the column headed "D," by the number of sec- 
onds, and add the product to the preceding result, for the 
sine or tangent, and subtract it therefrom for the cosine or 



cotangent. 



EXAMPLES. 



1. Find the logarithmic sine of 40° 26' 28". 

OPERATION. 

log sin 40° 26' 9.811952 

Tabular difference 2.47 

No. of seconds 28 

Product • • • 69.16 to be added • • 69 

log sin 40° 26' 28" #9.812021 






PLANE TRIGONOMETRY. 25 

The same rule is followed for decimal j^artSj as in Art. 12. 

2. Find the logarithmic cosine of 53° 40' 40". 

OPEEATION. 

log cos 53° 40' 9.772675 

Tabular difference 2.86 

No. of seconds 40 

Product • • • 114.40 to be subtracted 114 

log cos 53° 40' 40" 9.772561 

If the arc is greater than 90°, we find the required 
function of its supplement (Arts. 26 and 28). 



3. Find the logarithmic tangent of 118° 18' 25". 



OPERATION. 

180° 

Given arc 118° 18' 25" 

Supplement 61° 41' 35" 

log tan 01° 41' 10.268556 

Tabular difference 5.04 
No. of seconds 35 

Product • • • 176.40 to be added • 176 

log tan 118° 18' 25" 10.268732 



4. Find the logarithmic sine of 32° 18' 35". 

Ans. 9.727945. 

5. Find the logarithmic cosine of 95° 18' 24". 

Ans. 8.000080. 

6. Find the logarithmic cotangent of 125° 23' 50". 

Ans. 9.851619. 



26 PLANE TRIGONOMETRY. 

To find the arc corresponding to any logarithmic fimction, 

36. This is done by reversing the preceding rule : 
Look in the proper column of the table for the given log- 
arithm ; if it is found there, the degrees are to be taken 
from the top or bottom, and the minutes from the left or 
right hand column, as the case may be. If the given log- 
arithm is not found in the table, then find the next less 
logarithm, and take from the table the corresponding degrees 
and minutes, and set them aside. Subtract the logarithm 
found in the table, from the given logarithm, and divide the 
remainder by the corresponding tabular difference. The quo- 
tient will be seconds, which must be added to the degrees 
and minutes set aside, in the case of a sine or tangent, and 
subtracted, in the case of a cosine or a cotangent. 

EXAMPLES. 

1. Find the arc corresponding to the logarithmic 
sine 9.422248. 

OPERATION. 



Given logarithm • • • 9.422248 

Next less in table • • • 9.421857 • • • 15° 19' 

Tabular difference 7.68) 391.00(51", to be added. 

Hence, the required arc is 15° 19' 51". 

2. Find the arc corresponding to the logarithmic 
cosine 9.427485. 

OPERATION. 

Given logarithm . . • 9.427485 

Next less in table • • 9.427354 • . . 74° 29'. 

Tabular difference 7.58 ) 131.00 ( 17 , to be subt. 

Hence, the required arc is 74° 28' 43". 



PLANE TRIGONOMETRY. 



27 



3. Find the arc corresponding to the logarithmic 
sine 9.8S0054. Ans. 49* 20' 50". 

4. Find the arc corresponding to the logarithmic 
cotangent 10.0086S8. Ans. 44° 25' 37". 

5. Find the arc corresponding to the logarithmic 
cosine 9.944599. Ans. 28° 19' 45". 






SOLUTION OF RIGHT-ANGLED TRIANGLES. 

37. In what follows, we shall designate the three angles 
of every triangle, by the capital letters A, B, and C, A 
denoting the right angle ; and the sides lying opposite the 
angles, by the corresponding small letters a, b, and c. 
Since the order in which these letters are placed may be 
changed, it follows that whatever is proved with the letters 
placed in any given order, will be equally true when the 
letters are correspondingly placed in any other order. 

Let CAB represent any triangle, 
right-angled at A. "With C as a 
centre, and a radius CB, equal to 1, 
describe the arc BG, and draw GF 
and BE perpendicular to CA : then 

will FG be the sine of the angle C, CF will be its 
cosine, and DE its tangent. 

Since the three triangles CFG, CDF, and CAB are 
similar (B. IV., P. XVIII.), we may write the proper 
tions, 




CB : CG : 
CB : CG : 
CA : CD : 



AB : FG, or, a 
CA : CF, or, a 
AB : BE, or, b 



1 

] 
1 



c : sin C 
b : cosC 
c : tan C ; 



28 PLANE TRIGONOMETRY, 

hence, we have (B. II., P. I.), 



o = a sin C • • 


' • (1.) 




sin C = — , • « 
a 


■ (*•) 


b = a cos (7 • < 


• (2.) 




cos C = — , • 
a 


• • (5.) 


c = b tan G • - 


• ■ («0j 




tan (7 = -—■ , • 
^ o 


• • (6.) 



Translating these formulas into ordinary language, we have 
the following 

PEIXCIPLES. 

1. The perpendicular of any right-angled triangle is equat 
to the hypotlienuse into the sine of the angle at the base. 

2. The base is equal to the hypothenuse into the cosine 
of the angle at the base. 

3. The perpendicular is equal to the base into the tan- 
gent of the angle at the base. 

4. The sine of the angle at the base is equal to the 
perpendicidar divided by the hypothenuse. 

5. The cosine of the angle at the base is equal to the 
base divided by the hypothenuse. 

6. The tangent of the angle at the base is equal to the 
perpendicular divided by the base. 

Either side about the right angle may be regarded as the 
base ; in which case, the other is to be regarded as the 
perpendicular. We see, then, that the above principles are 
sufficient for the solution of every case of right-angled tri- 
angles. When the table of logarithmic sines is used, in the 
solution, Formulas ( 1 ) to ( 6 ) must be made homogeneous, 
by substituting for sin C, cos (7, and tan (7, respectively, 





R 




a 


cos 


G 


i 


R 




b 


tan 


G 



■ (»•) 


sin G 


Re 

— — - • 


• ■ (io.) 


• (8.) 


cos G 


2» 

a 


• (11.) 


• (o.) 


tan G 


Ro 

— — • 


• • (12.) 



PLANE TRIGONOMETRY. 29 

gin G co3 G , tan G „ , . . 

— — — , — — - , and — p— , R being equal to 

10,000,000,000, as explained in Art. 30. 

Making these changes, and reducing, -we have, 

a sin C 
c = 



b = 



C = R 



In applying these formulas, four cases may arise ' 

CASE I. 

Given the hypothenuse and one of the acute angles, to find 

the remaining parts. 

38. The other acute angle may be found by subtracting 
the given one from 90° (Art. 23). 

The sides about the right angle may 
be found by Formulas ( 7 ) and ( 8 ). 

EXAMPLES. 

1. Given a — 749, and G = 47° 03' 10" ; required 
i?, b, and c. 

OPERATION. 

R — 90° — 47° 03' 10" = 42° 50' 50". 

Applying logarithms to Formula (7), remembering that the 
logarithm of R is equal to 10, we have, 

log c — log a + log sin G — 10 ; 

log a (749) .... 2.874482 

log sin C (-1-7° 03' 10") • 9.804501 

log c 2.738983 .'. C = 548.255. 

19 




30 PLANE TRIGONOMETRY. 

Applying logarithms to Formula (8), we have, 

log b — log a -f- log cos G — 10 ; 

log a (749) - • • ■ 2.874481 

log cos G (47° 03' 10") • 9.8333 54 

log b 2.707835 .'. b = 510.31 

Ans. B = 42° 56' 50", b = 510.31, and c = 548.255 

2. Given a = 439, and B = 27° 38' 50", to find 
C, b, and c. 

OPERATION. 

C — 90° — 27° 38' 50" — 62° 21' 10" ; 

log a • (439) .... 2.642465 
log sin G (62° 21' 10") ■ 9.947346 

loo- c 2.589811 .'. c = 388.875 • 

log <b • (439) .... 2.642465 
log cos G (62° 21' 10") • 9.666543 

loo- b 2.309008 .'. b — 203.708. 

Ans. G = 62° 21' 10", b = 203.708, and c = 388.875. 

3. Given a = 125.7 yds., and B = 75° 12', to find 
die other parts. 

Ans. G = 14° 48', 5 = 121.53 yds., and c = 32.11 yds 

4. Given a = 325 ft., and (7 = 27° 34', to find th 
other parts. 

Ans. B = $2° 26', c = 150.4 ft., and b = 288.1 ft. 



PLAXE TRIGONOMETRY, 81 

CASE II. 

Given one of the sides about the right angle a?id one of 
the acute angles, to find the remaining parts. 

39. The other acute angle may be found by subtracting 
the given one from 90°. 

The 'hypothenuse may be found by Formula (7), and 
the unknown side about the right angle, by Formula (8). 

EXAMPLES. 

1. Given c = 56.293, and C = 54° 27' 39", to find £, 
a, and b. 

OPERATION. 

J3 = 90° — 54° 27' 39" = 35° 32' 21". 

Applying logarithms to Formula (7), we have, 

log c = log a + log sin C — 10 ; whence, 

log a = log c + 10 - log sin C = log c + (a, c.) log sin C; 

log c (5G.293) . . . 1.750454 

(a. c.) log sin C (54° 27' 39") • 0.089527 

log a 1.839981 .-. a = 69.18. 

Applying logarithms to Formula (8), we have, 

log b = log a -f log cos C — 10 ; 

log a (G9.1S) .... 1.8399S1 

log cos O (54° 27' 39") • • 9.7G4370 

lo S o 1.604351 .-. 5 = 40.2114. 

Am. B = 35° 32' 21", a i= 69.18, and b = 40.2114. 



32 PLANE TRIGONOMETRY. 

2. Given c = 358, and B = 28° 47', to find (7, a, 
and d 

OPERATION. 

G ■= 90° - 2S° 47' = 61° 13'. 

We have, as before, 

log a = log c + (a. c.) log sin (7, 

and, log 5 = log a + log cos (7 — 10 ; 

log c (358) • • • 2.553883 

(a. c.) log sin G (61° 13') - • 0.057 274 

log a 2.611157 .'. a = 408.466; 

log a (313.776) - "- 2.611157 

log cos G (61° 13') • • 9.682595 

loo- b 2.293752 .\ b = 196.676. 

'Ans. G = 61° 13', a = 408.466, and b = 196.676. 

3. Given b = 152.67 yds., and C = 50° 18' 32", to 
find the other parts. 

Ans. B = 39° 41' 28", c = 183.95, and a = 239.05. 

4. Given c = 379.628, and G == 39° 26' 16", to find 
i?, a, and b. 

Ans. B = 50° 33' 44", a = 597.613, and b - 461.55, 

CASE III. 

Given the two sides about the right angle, to find the re 

maining parts. 

40. The angle at the base may be found by Formula 
(12), and the solution may be completed as in Case II. 



PLANE TRIGONOMETRY. S3 

EXAMPLES. 

1. Given b = 26, and e '= 15, to find G, B, and a. 

OPERATION. 

Applying logarithms to Formula (12), we have, 

« 

•'log tan C = log c + 10 — log 5 = log c + (a. c) log b ; 

log c (15) .... 1.176091 
(a. c.) log b (26) .... 8.585027 

log tan G • • • 9.761118 .\ G = 29° 58' 54" ; 

j? _ 90 o _ (7 - 60° 01 r 06". 
As in Case II., log a = log c + (a. c.) log sin G ; 

log a • • (15) • • 1.176091 
(a. c.) log sin G (29° 58' 54") 0.301 271 

loo- « 1.477362 .*. a — 30.017. 

Am. C = 29° 58' 54", B = 60° 01' 06", and a = 30.017. 

4 + 

2. Given b — 1052 yds., and c = 347.21 yds.,, to find 
i?, (7, and a. 

B = 71° 44' 05", 6' = 18° 15' 55", and a = 1108.05 yds. 






3. Given b = 122.416, and c = 118.297, to find i>\ 

C T , and a. 

B = 45° 58' 50", G = 44° 1' 10", and a = 170.235 

4. Given ft = 103, and c = 101, to find I?, (7 

and a. 

B = 45° 33' 42", (7 = 44° 26' 18", and a = 144.250. 



34 PLANE TRIGONOMETRY. 



CASE IV. 

Given the Jiypothenuse and either side about the right angle^ 

to find the remaining parts, 

41. The angle at the base may be found by one of 
Formulas (10) and (11), and the remaining side may then 
be found by one of Formulas (7) and (8). 

EXAMPLES. 

1. Given a = 2391.76, and b == 385.7, to find B> 
and c. 

operation. 

Applying logarithms to Formula (11), we have, 

log cos G = log b + 10 - log a = log b + (a. c.) log a ; 

log b (385.7) • • • 2.586250 
(a. c.) log a (2391.76) • • 6.621282 

loo- cos G • • v 9.207532 .'. G = 80° 43' 11"; 



£ _ 90 o _ 80° 43' 11" = 9° 16' 49". 

From Formula ( 7 ), we have, 

log c = log a + log sin G — 10 ; 

log a (2391.76) - 3.378718 

log sin G (80° 43' 11") 9.994278 

log c 3.372996 .'. C = 2360.45. 

Ans. B = 9° 16' 49", C = 80° 43' 11", and c = 2360.45. 



PLANE TRIGONOMETRY. 35 

2. Given a = 127.174 yds., and c = 125.7 yds., to 
find B > C, and b. 

OPERATION. 

From Formula ( 10 ), we have, 

log sin G = log c + 10 — log a = log c + (a. o.) log a ; 

log C (125.7) • • • 2.099335 
(a. c.) log a (127.174) - - 7.895602 

log sine- • • • 9.994937 .\ <7 = 81° 16' 6" ; 

B — 90° — 81° 16' 6" — 8° 43'- 54". 
From Formula (8), Ave have, 

log b = log a + log cos C — 10 ; 

log a (127.174) • 2.104398 

log cos G (81° 16' 6") • 9.181292 

log b 1.285690 .'. b — 19.3. 

Ans. B = 8° 43' 54", G = 81° 16' 6", and b = 19.3 yds. 

3. Given a = 100, and 6 = 60, to find B, C> and ft 
'-4n*. I? = 36° 52' 11", (7 = 53° 7' 49", and c = 80. 

4. Given a = 19.209, and c = 15, to find 7>, (7, 

and b. 

Ans. B = 38° 39 30" (7 = 51° 20' 30", b = 12. 






36 



PLANE TRIGONOMETRY. 



SOLUTION OF OBLIQUE-ANGLED TEIANGLES. 

42, In the solution of oblique-angled triangles, four cases 
may arise. We shall discuss these cases in order. 



CASE I. 

Given one side and two angles, to determine the remaining 

parts, 

43. Let ABC represent any 
oblique-angled triangle. From the 
vertex (7, draw CD perpendicular 
to the base, formiug two right- 
angled triangles A CD and BCD. 
Assume the notation of the figure. 

From Formula ( 1 ), we have, 




CD — b sin A, 



and 



CD = a sin B ; 



Equating these two values, we have, , 

b sin A = a sin B ; 

whence (B. II., P. II.), 

a : b : : sin A : sin B. . . ( 13.) 

Since a and b are any two sides, and A and B the 
angles lying opposite to them, we have the following princi- 
ple : 

The sides of a plane triangle are proportional to the 
Bines of the opposite angles. 

It is to be observed that Formula (13) is true for any 
value of the radius. Hence, to solve a triangle, when a side 
and two angles are given: 



PLANE TRIGONOMETRY. 37 

First find the third angle, by subtracting the sum of the 
given angles from 180° ; then find each of the required sides 
by means of the principle just demonstrated. 

EXAMPLES. 

1. Given B = 58° 07', C = 22° 37', and a = 40S, to 
find A, b, and c. 

OFEEATION. 

J? 58° 07' 

C 22 ° 37' 

A . . . 180° - 80° 44' = 99° 16'. 

To find b, write the proportion, 

sin A : sin B : : a : b ; 

that is, the sine of the angle opposite the given side, is to 
the sine of the angle opposite the required side, as the given 
side is to the required side. 

Applying logarithms, and reducing, we have, 

log b = log a + log sin B + (a. c) log sin A — 10 ; 

log a • • (408) - • • • 2.610G60 

log sin B (58° 07') - • • 9.928972 

(a. c) log sin A (99° 16') • • • 0.005 705 

\ orr o 2.545337 .'. 5 = 351.024. 

O t 

In like manner, 

log c = log a + log sin C + (a. c.) log sin A - 10 ; 

log a • • (408) • • • 2.G10GG0 

lorr sin C (22° ?V) • • • 9.5849G8 

(a, c.) log sin A (99° J6') ■ • • 0-005705 

W c 2.201333 . . c = 158.97G. 

O — ■ 

Ans. A = 99° 16', b = 351.024, and c = 158.970. 



38 PLANE TRIGONOMETRY. 

2. Given A = 38° 25', B — 57° 42', and c = 400, 
to find C, a, and 5. 

^s. (7 = 83° 53', a = 249.974, b = 340.04. 

3. Given A = 15° 19' 51", (7 == 72° 44' 05", and 
c = 250.4 yds, to find B, a, and b. 

Ans. B — 91° 56' 04", a = 69.328 yds., b — 262.066 yds. 

4. Given B = 51° 15' 35", C — 37° 21' 25", and 
a — 305.296 ft., to find A, b, and c. 

Ans. A — 91° 23', b — 238.1978 ft., c = 185.3 ft. 



CASE II. 

Given two sides and an angle opposite one of them, to find 

the remaining parts. 

44. The solution, in this case, is commenced by finding 
a second angle by means of Formula (13), after which we 
may proceed as in Case I. ; or, the solution may be com- 
pleted by a continued application of Formula (13). 

EXAMPLES, 

1. Given A = 22° 37', b = 216, and a = 117, to 
find B, (7, and c. 

From Formula (13), we have, 

a : b : : sin A : sin B ; 

that is, the aide opposite the given angle, is to the side op- 
posite the required angle, as the sine of the given angle is 
to the sine of the required angle. 



PLANE TRIGONOMETRY. 



39 



Whence, by the application of logarithms, 

log sin B — log b + log sin A + (a. c.) log a — 10 ; 



log b • • (216) 
log sin .4 (22° 37') 
(a. c.) log a - - (117) 



log sin B 



2.334454 
9.584968 
7.931814 
9.851236 .-.JB 



: 45° 13' 55", 
and B' — 134° 46' 05". 




Hence, we find two values of B, which are supplements of 
each other, because the sine of any angle is equal to the 
sine of its supplement. This would seem to indicate that 
the problem admits of two solutions. It now remains to 
determine under what conditions there will be two solutions, 
one solution, or no solution. 

There may be two cases : the given angle may be acute, 

or it may be obtuse. 

First Case. Let ABC re- 
present the triangle, in which the 
ano-le A, and the sides a and 

b are given. From C let fall '** "* 

a perpendicular upon AB, pro- 
longed if necessary, and denote its length by p. We shall 
have, from Formula ( 1 ), Art. 37, 

p = b sin A ; 

from which the value of p may be computed. 

If a is intermediate in value between p and b, there 
will be two solutions. For, if with C as a centre, and n 
as a radius, an arc be described, it will cut the line AB 
in two points, B and B ', each of which being joined with 
C, will give a triangle which will conform to the conditions 
of the problem. 



40 



PLANE TRIGONOMETRY. 



In this case, the angles B' and B, of the two triangles 
AJB'C and ABC, will be supplements of each other. 

C 



If 



a 



p, there will be but 




one solution. For, in this case, 
the arc will be tangent to AB, 
the two points B and B' will 
unite, and there will be but a single triangle formed. 
In this case, the angle ABC will be equal to 90°. 

If a is greater than both p 
and b, there will also be but one 
solution. For, although the arc 
cuts AB in two points, and con- 
sequently gives two triangles, only 
one of them conforms to the con- 
ditions of the problem. 

In this case, the angle ABC will be less than A, and 
consequently acute. 



If a < p, there will be no 
solution. For, the arc can neither 
cut AB, nor be tangent to it. 





Second Case. When the given angle A is obtuse, the 
angle AB C will be acute ; the 
side a will be greater than b, 
and there will be but one solu- 
tion. 

In the example under considera- 
tion, there are two solutions, the 

first corresponding to B = 45° 13' 55", and the second to 
B' = 134° 40' 05". 







PLANE TRIGONOMETRY. 41 

In the first case, we have, 

A 22° 37' 

B 45° 13' 55" 

C 180° — C7° 50' 55" == 112° 09' 05". 



As in Case I., we have, 

log c — log b -f log sin C + (a. c.) log sin B — 10 ; 

log b • • (216) • • • 2.334454 

log sin C (112° 09' 05") • 9.966700 

(a c.) log sin B ( 45° 13' 55") • 0.146764 

lo" c - • 2.449918 .'. c = 281 785. 



o 



Ans. B = 45° 13' 55", C = 112° 09' 05", and c = 281.785. 

In the second case, we have, 

A 22° 37' 

B' 134° 4 6' 05" 

C 180° - 157° 2 3' 05" - 22° 36' 55"; 

and as before, 

log b • • (216) • • • 2.334454 

log sin C (22° 36' 55") • 9.5S4943 

(a. c.) log sin B (134° 46' 05") • 0.146764 

k>£ c 2.0681-61 .'. c = 116.993. 

Ans. B' = 134° 46' 05", C = 22° 36' 55", and c = 116.993. 

2. Given A = 32°, a = 40, and b = 50, to find 

2?, (7, and c. 

B =z 41° 28' 59", C = 106° 31' 01", c = 72.368. 
Ans. 

B = 138° 31' 01", C = 9° 28' 59", c = 12.436. 



42 



PLA^E TRIGONOMETRY. 



3.* Given A = 18° 52' 13'', a = 27.465 yds., and 
b = 13.189 yds., to find B, C, and c. 

Ans. B = 8° 56' 05", (7 = 152° 11' 42", c = 39.611 yds. 



4. Given ^ = 32° 15' 26", b = 176.21 ft., and 
a = 94.047 ft., to find B, C, and e. 

Ans. B = 90°, G = 57° 44' 34", c = 149.014 ft. 




cr--,._-^'F H 



case ni. 

Given two sides and their included angle, to find the re- 

maining parts. 

45. Let ABC represent any 
plane triangle, AB and AG any 
two sides, and A their included 
angle. With i as a centre, 
and AC, the shorter of the two 
sides, as a radius, describe a semi- 
circle meeting AB in I, and the prolongation of AB 
in E. Draw CI and i?(7, and through I draw ZZ7 
parallel to EC. 

Because ECI is an angle inscribed in a semicircle, it is 
a right angle (B. EX. P. XVIII., C. 2) ; and consequently, 
both CE and III are perpendicular to CI. The angle 
EAC being external to the triangle ABC, is equal to the 
sum of the opposite interior angles, that is, equal to C 
plus B ; the angle EA C being also external to the isos- 
celes triangle AIC, it is equal to twice the angle AIC : 
hence, twice the angle AIC is equal to C plus B, or, 

AIC = i(C + B). 



PLANE TRIGONOMETRY. 43 

The angle ICB is equal to AIG diminished by the angle 
IB C (B. I., P. XXV., C. 6) ; that is, 

ICH = i{C+JB)-B = i(G - B). 

From the two right-angled triangles ICE and ICH, we 
have (Formula 3, Art. 37), 

EC = IG tau^(<7 + .B), and IH = IG tan i ( (7 - 7?) ; 

hence, from the preceding equations, we have, after omitting 
the equal factor IG (B. IX, P. VII.), 

EG : IS : : tan i{C + B) : tan i[0 - B). 

The triangles ECB and ZZIZ£ being similar, their homo- 
Jogous sides are proportional ; and because EB is equal to 
AB + AC, and IB to AB — AG, we shall have the 
proportion, 

EG : IH :: A2? + ylC : .47? - AG, 

Combining the preceding proportions, and substituting for 
AB and AG their representatives c and b, we have, 

c + fl : c-b :: tanJ(tf+.Z?) : tani(ff-'JB) . . (14.) 

Hence, we have the following principle : 

7n any ^a?2e triangle, the sum of the sides including 
either angle, is to their difference, as the tangent of half 
the sum of the two other angles, is to the tangent of half 
their difference. 

The half sum of the angles may be found by subtracting 
the ghen angle from 180°, and dividing the remainder by 2 
the half difference may be found by means of the principle 
just demonstrated. Knowing the half sum and the half 



: i :, : z ::,:>::":z:::,7 



__ZL£T£i:-z ZZf ZTcZZZI LZZlf l- : 1" — - 1; 

z :t_ : :: ■_. _ - r — zz L zzi '_-.:* :z : = . : zz L 

: - ; - _ " _ - 1 : z : : i t : zz " _ - : — — SZ- 

z:z : ' : — I --'-'- - " — - " ' ~ - 

X. G:zz = 5411, * = i5H, ami J = 80% to 

£ . •_"*. Ill A. 

: : zz__z: 

c + * = 990; « — £ = «»; | F+J = IflS®* — «P) = 

:.z r_z T : -x = - : - ' - i:c -~- i r - - - 

lag (c -!- *) — IP ; 

log^ — J) . - (») 1^1243 

log ta» - - -3) (iop) l®ue£®- 

(*. cl) log - 3 - - (990) " -^- - 

* ■ ." - 1 '- "---'■: t ~~- = ?'-!!*: 



= - 5 s 11 = »■ :: - = KP-rii' = ::»«_ 

F'zz L~ — T~~, 1! ~~ t zz.~5. 

log « = kc - kg a Jl - (sl e.) log szz - 10 ; 



log sd A T} - - 9-9SSS51 



- -: = f ; : :: . : = :v. :■?■: 



PLAN 



RIGOXOMETRT. 



.5 



2. Gzzz ..? =:•::•:■ 7 1=., : = &-:: 71=., -: j. = :;s : :~', 

to rii I and a. 

.; , i? = :s : :;'::. '."" = : : : :-. :: , ■:: = :-::: 7.:s. 
n .-! = :s.::? 7" . :• = ~.:z; 7.1$.. : 1 1 

C — ^; : 1: 28", :: zzz A. B. uzl ■:■. 

A w. -i = ::: : 34 :: . B = ■::'■ ;-::-. c= 14.42c 7:, 



4. G 

.5. J = ■: ". : . 



4 - 7 7 Is, 6 = :39.3 ydsL, and 

: : find A. B, and e. 

'39", B= 32°42'33", e = 534.66 yda. 



5. Given = 1 : : z 5 = 11.9613 : zJ 

£7 = 60® 4 . ::■ zzl -4. B. ::.'. ■:. 

An*. A = ' A' 10", J? = 43" 12 14 ', c = 15.22 ft. 

6. Given a = 3754, '- = 32 " .628, and C=': : : 17. 
find -1 J. and 

An*. A = 0: . - \ B = :^ : M 18", e = £428.512. 



~Z IT. 



fee* *A« Mree «Mfe* of a triangle, to find the remain i 

parti* 



46. Let -17 represent any 
plane triangle, of which B 
the k 1 aw -<4I> per- 

pendicnlar to the base, dividing it 
into two sesnnents b 1j ar. 1 J 




• The angles may be found by Formula (JD or 3 rzvax. Pages 

109, and 110, Mensuration. 

-. 



4:6 



PLANE TRIGONOMETRY. 



From the right-angled triangles BAB and GAB, we 

have, 

AB 2 = AB 2 - BB 2 , and AD 2 = AG 2 - DC 2 ; 



Equating these values of AB , we have, 



AM 2 - BB 2 = AG 2 - BG 2 ; 
whence, by transposition, 




■AG* - AB" = BG' - BB\ 

Factoring each member, we have, 

(AG + AB) (AG-AB) = {BG + BB) (BC-BB). 

Converting this equation into a proportion (B. II., P. II.), 
we have, 
BG + BB : AG + AB :: AG - AB : BG-BB; 



or, denoting the segments by s and 
<of the triangle by a, b, and c, 



s', and the sides 



s + s' 



b + c 



s — s 



t . 



(15.) 



that is, if in any plane triangle, a line be drawn from the 
vertex of the vertical angle perpendicular to the base, divid- 
ing it into two segments ; then, 

The sum of the two segments, or the whole base, is to 
ithe sum of the tioo other sides, as the difference of these 
.sides is to the difference of the segments. 

The half difference added to the half sum, gives the 
greater, and the half difference subtracted from the half sum 
mves the less segment* We shall then have two right- 
angled triangles, in each of which we know the hypothenuse 
and the base ; hence, the angles of these triangles may be 
found, and consequently, those of the given triangle. 



PLANE TRIGONOMETRY. 



m 



EXAMPLES. 

1. Given a = 40, b = 34, and e = 25, to find 
#, and C. ' 

OPERATION. 

Applying logarithms to Formula (15), we have, 



log (s — s') — log (b -f c) + log (5 — c) + (a. c.) log (s -f s')— 10; 



log(5+c) - (59) 

log {b - (?) • (9) 

(a. c.) log (s + s') • (40) 

log (s — s') • • 



. 1.770852 

. 0.954243 

• 8.397940 

• 1.123035 .\ s — s' = 13.27S. 



5 =z l(s + s') + i(s - s') = 26.6375 
5' = i(s + s') — |(s — s') = 13.3625 
From Formula (11), we find, 

log cos C = log 5 + (a. c.) log b .'. C = 38° 25' 20"j and 

log cos I? = log s' + (a. c.) log c .\ i? = 57° 41' 25" 

9G° 06' 45" 



A = 180° - 96° 06' 45" = 83° 53' 15". 



2. Given a = 6, 5 = 5, and c = 4, to find A, 
By and (7. 

-4rcs. A = 82° 49' 09", JO = 55° 46' 16", C = 41° 24' 35" 



3. Giv r en a = 71.2 yds., b = 64.8 yds., and c = 37.4 
yds., to find A 9 £, and C. 

Am. A = 83° 44' 32", i? = 64° 46' 50", C = 31° 28' 30", 



48 



PLANE TRIGONOMETRY. 



PEOBLEMS. 

1. Knowing the distance AB, 
equal to 600 yards, and the angles 
BAG = 57° 35', ABC = 64° 51', 
find the two distances AC and 
BO. 




Ans. AC — 643.49 yds., BC = 600.11 yds. 



2. At what horizontal distance from a column, 200 feet 

high, will it subtend an angle of 31° 17' 12" ? 

Ans. 329.114 ft. 



3. Required the height 
of a hill B above a hor- 
izontal plane AB, the dis- ^--- * 
tance between A and B j |^ | 




C 



being equal to 975 yards, 
and the angles of elevation at A and B being respect- 
ively 15° 36' and 27° 29'. Ans. BC = 587.61 yds. 



4. The distances J.C and BC 
are found by measurement to be, res- 
pectively, 588 feet and 672 feet, and 
their included angle 55° 40'. Requir- 
ed the distance AB. 

Ans. 592.967 ft. 



5. Being on a horizontal plane, and wanting to ascertain 
the height of a tower, standing on the top of an inaccessible 
hill, there were measured, the angle of elevation of the top 
of the hill 40°, and of the top of the tower 51° ; then 
rneasiiiii^ III st direct line 180 feet farther from the hill, the 




PL AXE TRIGONOMETRY 



49 



angle of elevation of the top of the tower was 33° 45' ; 
required the height of the tower. Ans. 83.998 ft. 

6. Wanting to know the horizontal distance between two 
inaccessible objects E and TF, the 
following measurements were made : 



AB — 536 yards 

BAW = 40° 16' 

WAE = 57° 40' 

ABE = 42° 22' 

EBW .= n° 07'. 




Required the distance EW. 



Ans. 939.634 yds. 




7. Wanting to know the 
horizontal distance between 
two inaccessible objects A 
and 7?, and not finding any 
station from which both of 
them could be seen, two 
points C and Z>, were chosen 
at a distance from each other 

equal to 200 yards ; from the former of these points, A 
could be seen, and from the latter, B ; and at each of the 
points C and D, a staff was set up. From C, a dis- 
tance CF was measured, not in the direction DC, equal 
to 200 yards, and from Z>, a distance BE, equal to 200 
yards, and the following angles taken : 



AFC = 83° 00', 
BBC = 156° 25', 



BBE = 54° 30', ACB = 53° 30' 
ACF = 54° 31', BED = S8° 30' 



Required the distance All. 



Ans. 345 4 67 yds. 



so 



PLANE TRIGONOMETRY. 



8. The distances AB, AC, and 
BC, between the points A, B, and 
C, are known ; viz. : AB — 800 yds., 
AC = 600 yds., and BC = 400 yds. 
From a fourth point jP, the angles 
ABC and BBC are measured : 



viz. 



JLPC = 33° 45', 




and BBC= 22° 30'. P~ 

Required the distances AB, BB, and <7P. 

r^p = 

Ans. \ BP =z 



710.193 .yds, 
934.291 yds. 



^ CP = 1042.522 yds. 

This problem is used in locating the position of buoys in 
maritime surveying, as follows. Three points A, B, and 
C, on shore are known in position. The surveyor stationed 
at a buoy P, measures the angles APC and BBC. The 
distances AP, BB, and CP, are then found as follows : 

Suppose the circumference of a circle to be described 
through the points A, P, and P. Draw CP, cutting 
the circumference in P, and draw the lines BB and BA. 

The angles CBB and BAB, being inscribed in the 
same segment, are equal (B. III., P. XVIII., C. 1) ; for a 
like reason, the angles CBA and BBA are equal : hence, 
in the triangle ABB, we know two angles and one side ; 
we may, therefore, find the side BB. In the triangle ACB, 
we know the three sides, and we may compute the angle P. 
Subtracting from this the angle BBA, we have the angle 
BBC. Now, in the triangle BBC, we have two sides 
and their included angle, and we can find the angle BCB. 
Finally, in the triangle CBB, we have two angles and one 
side, from which data we can find CP and BP. In like 
manner, we can find AB, 



ANALYTICAL TRIGONOMETRY. 



47. Analytical Trigonometry is that branch of Mathe- 
matics which treats of the general properties and relations 
of trigonometrical functions. 



DEFINITIONS AND GENERAL PRINCIPLES. 




48. Let ABCB represent a cir- 
cle whose radius is 1, and suppose 
its circumference to be divided into 
four equal parts, by the diameters 
AC and BB, drawn perpendicular to 
each other. The horizontal diameter 
A C, is called the initial diameter ; 
the vertical diameter BB, is called 

the secondary diameter ; the point A, from which arcs are 
usually reckoned, is called the origin of arcs, and the point 
B, 90° distant, is called the secondary origin. Arcs esti- 
mated from A, around towards B, that is, in a direction 
contrary to that of the motion of the hands of a watch, are 
considered positive ; consequently, those reckoned in a con- 
trary direction must be regarded as negative. 

The arc AJB, is called the first quadrant; the arc BO, 
the second quadrant; the arc CB, the third quadrant; 
and the arc CA, the fourth quadrant. The point at which 



52 



ANALYTICAL 



an arcs terminates, is called its extremity, and an arc is said 
to be in that quadrant in which its extremity is situated. 
Thus, the arc AM is in the first 
quadrant, the arc AM' in the sec- 
ond, the arc AM" in the third, 
and the arc AM'" in the fourth. 




tn 



49. The complement of an arc 
has been defined to be the differ- 
ence between that arc and 90° (Art. 
23) ; geometrically considered, the 

complement of an arc is the arc included between the extremity 
of the arc and the secondary origin. Thus, MB is the 
complement of AM ; M'B, the complement of AM' ; 
M"B, the complement of AM", and so on. When the 
arc is greater than a quadrant, the complement is negative, 
according to the conventional principle agreed upon (Art. 48). 

The supplement of an arc has been defined to be the 
difference between that arc and 180° (Art. 24) ; geometrically 
considered, it is the arc included between the extremity of 
the arc and the left hand extremity of the initial diameter. 
Thus, MQ is the supplement of AM, and M"0 the sup- 
plement of AM'. The supplement is negative, when the 
arc is greater than two quadrants. 



50. The sine of an arc is 
the distance from the initial 
diameter to the extremity of the 
arc. Thus, PM is the sine 
of AM, and P"M" is the 
sine of the arc AM". The 
term distance, is used in the 
sense of shortest or perpendicu- 
lar distance. 




TRIGONOMETRY. 53 

51. The cosine of an are is the distance from the sec- 
ondary diameter to the extremity of the arc : thus, NM 
is the cosine of AM, and NM 1 is the cosine of AM'. 

The cosine may be measured on the initial diameter : 
thus, OP is equal to the cosine of AM, and OP' to the 
cosine of AM'. 

52. The versed-sine of an arc is the distance from the 
sine to the origin of arcs : thus, PA is the versed-sine of 
AM, and P'A is the versed-sine of AM'. 

53. The co-versed-sine of an arc is the distance from 
the cosine to the secondary origin : thus, KB is the co- 
versed-sine of AM, and JS T "B is the co-versed-sine of AM". 

54. The tangent of an arc is that part of a perpen- 
dicular to the initial diameter, at the origin of arcs, in- 
cluded oeticeen the origin and the prolongation of the diam- 
eter through the extremity of the arc : thus, AT is the 
tangent of AM, or of AM", and AT" is the tangent 
of AM\ or of AM'". 

55. The cotangent of an arc is that part of a perpen- 
dicular to the secondary diameter, at the secondary origin, 
included between the secondary origin and the prolongation 
of the diameter through the extremity of the arc : thus, 
BT' is the cotangent of AM, or of AM", and BT n 
is the cotangent of AM', or of AM'". 

56. The secant of an arc is the distance from the cen- 
tre of the arc to the extremity of the tangent : thus, OT 
is the secant of AM, or of AM", and OT'" is the so 
cant of AM', or of AM'". 

57. The cosecant of an arc is the disf"»ce from the 



54 



ANALYTICAL 



centre of the arc to the extremity of the cotangent : thus, 
OT' is the cosecant of AM, or of AM", and OT" is 
the cosecant of AM% or of AM'". 

The term co, in combination, is equivalent to complement 
of ; thus, the cosine of an arc is the same as the sine of 
the complement of that arc, the cotangent is the same as 
the tangent of the complement, and so on. 

The eight trigonometrical functions above denned are also 
called circular functions. 



RULES FOE DETERMINING- THE ALGEBRAIC SIGNS OF CIRCULAR 

FUNCTIONS. 

58. All distances estimated upwards are regarded as pos- 
itive ; consequently, all distances estimated downwards must 
be considered negative. 

Thus, AT, PM, WB, P'M% 
are positive, and AT"', P'"M'", 



P"M'\ &c. 



are negative. 




M T'" 



All distances estimated towards 
the right are regarded as positive ; 
consequently, all distances estimat- 
ed towards the left must be con- 
sidered negative. 

Thus, NM, BT', PA, &c, 
are positive, and JST'31', BT", &c, are negative. 



All distances estimated from the centre in a direction to 
towards the extremity of the arc are regarded as positive ; 
consequently, all distances estimated in a direction from the 
second extremity of the arc must be considered negative. 

Thus, OT, regarded as the secant of AM, is estimated 
in a direction towards M, and is positive ; but OT, re- 



TRIGONOMETRY. 55 

garded as the secant of AM ", is estimated in a direction 
from 31", and is negative. 

These conventional rules, enable us at once to give the 
proper sign to any function of an arc in any quadrant. 

59. In accordance with the above rules, and the defini- 
tions of the circular functions, we have the following princi- 
ples : 

The sine is positive in the first and second quadrants, 

and negative in the third and fourth. 

The cosine is positive in the first and fourth quadrants, 
and negative in the second and third. 

The versed-sine and the co-versed-sine are always positive. 

The tangent and cotangent are positive in the first and 
third quadrants, and negative in the second and fourth. 

The secant is positive in the first and fourth quadrants, 
and negative in the second and third. 

The cosecant is positive in the first and second quadrants, 
and negative in the third and fourth. 

LIMITING VALUES OF THE CIRCULAR FUNCTIONS. 

60. The limiting values of the circular functions are those 
values which they have at the beginning and end of the 
different quadrants. Their numerical values are discovered 
by following them as the arc increases from 0° around to 
300°, and so on around through 450°, 540°, &c The signs 
of these values are determined by the principle, that the sign 
of a varying magnitude up to the limit, is the sign at the 
limit. For illustration, let us examine the limiting values of 
the sine and tangent. 



56 ANALYTICAL 

If we suppose the arc to be 0, the sine will be ; as 
the arc increases, the sine increases until the arc becomes 
equal to 90°, when the sine becomes equal to + 1, which is 
its greatest possible value ; as the arc increases from 90°, 
the sine goes on diminishing until the arc becomes equal to 
180°, when the sine becomes equal to + ; as the arc 
increases from 180°, the sine becomes negative, and goes on 
increasing numerically, but decreasing algebraically, until the 
arc becomes equal to 270°, when the sine becomes equal to 
— 1, which is its least algebraical value; as the arc increases 
from 270°, the sine goes on decreasing numerically, but in- 
creasing algebraically, until the arc becomes 360°, when the 
sine becomes equal to — 0. It is — 0, for this value of 
the arc, in accordance with the principle of limits. 

The tamrent is when the arc is 0, and increases till 
the arc becomes 90°, when the tangent is + co ; in passing 
through 90°, the tangent changes from + co to — co , and 
as the arc increases the tangent decreases, numerically, but 
increases algebraically, till the arc becomes equal to 180°, 
when the tangent becomes equal to — ; from 180° to 
270°, the tangent is again positive, and at 270° it becomes 
equal to + co ; from 270° to 360°, the tangent is again 
negative, and at 360° it becomes equal to — 0. 

If we still suppose the arc to increase after reaching 3G0°, 
the functions will again go through the same changes, that 
is, the functions of an arc are the same as the functions 
that are increased by 360°, 720° &c. 

By discussing the limiting values of all the circular funo 
tions we are enabled to form the following table : 



TRIGONOMETRY. 



57 



TABLE I. 



Arc = 


= 




Arc = 90°. 


Arc = 180°. 


Arc = 2*70°. 


Arc = 3G0°. 


Bin 


~ 





sin = 1 


sin = 


sin = — 1 


sin =—0 


cos 


= 


1 


cos = 


cos = — 1 


cos —-0 


cos = 1 


v-sin 


= 





v-sin = 1 


v-sin = 2 


v-sin = 1 


v-sin — 


co-v-sin 


= 


1 


co-y-sin = 


co-v-sin = 1 


co-v-sin = 2 


c-v-sin = 1 


tan 


— 





tan = co 


tan = — 


tan =. co 


tan =—0 


cot 


= 


CO 


cot = 


cot = — CO 


cot == 


cot = — CO 


sec 


— 


1 


sec = co 


sec =— 1 


sec = — co 


sec = 1 


coscc 


^z. 


CO 


cosec = 1 


cosec = co 


cosec = — 1 


cosec = — co 



RELATIONS BETWEEN THE CIRCULAR FUNCTIONS OF ANY ARC. 



61. Let AM represent any arc de- 
noted by a. Draw the lines as repre- 
sented in the figure. Then we shall 
have, "by definition 



OA = 1 ; JPM = OJST = sin a ; 

OP = cos a ; JPA = ver-sin a ; 

co-ver-sin a ; AT = tan a ; 




031 

NM 

NB 

BT' — cot a ; OT = sec a ; and OT' = cosec a. 



From the right-angled triangle OPM, we have, 



xmi 



P3f + OP = OM* , 



or. 



sin 2 a + cos 2 a = 1. . (1., 



The symbols sin 2 a, cos 2 a, &c, denote the square of the 
sine of «, the square of the cosine of a, &c. 
From Formula ( 1 ) we have, by transposition, 



in 2 a = 1 — cos 2 a . ( 2 ) ; and cos 2 a = 1 — sin 2 a. . ( 3.) 



Bin 



58 ANALYTICAL 

We have, from the figure, 
PA = OA - OP, 
or, ver-sin a = 1 — cos a. . . ^4.) 

and, JSTP = OP - ON, 

or, co-ver-sin a = 1 — sin <x. . . (5.) 



From the similar triangles OAT and OPM, we have, 

OP : PM : : OJ. : AT, or, cos a : sin a : : 1 : tan a; 

sin a , . 

whence, tana = — • ...... (6.) 

From the similar triangles OJSTM and 0J22 7 ', we have, 
OJST : NM : : OP : -B2 7 ', or, sin a : cos a : : 1 : cot a ; 

cos a , w x 

whence, cot a = ^-^ V'-J 

Multiplying (6) and (T), member by member, we have, 

tan a cot a = 1 ; (8-) 

whence, by division, 

tana = ^ ; • (9) and cota= f^' ' (10,) 

From the similar triangles OPIf and OAT, we have, 
OP : OM : : OA : OT, or, cos a : 1 : : 1 : sec a 
whence, »eo a = — • ("•) 



TRIGONOMETRY. 



59 



From the similar triangles ONM and OBT\ we have, 



ON" : OM : : OB : OT\ or, sin a : 1 : : 1 : co-sec a; 

1 



whence, 



% 
co-sec a 



sm a 



(12.) 



From the right-angled triangle OAT, we have, 



OT 2 = OA 2 -f- AT 5 or, sec 2 « = 1 -f tan 2 a. . (13.) 

From the right-angled triangle OBT\ we have, 
OT' 2 — OB 1 + BT' 2 \ or, co-sec 2 a = 1 + cot 2 a. . (14.) 

It is to be observed that Formulas (5), (7), (12), and 
(14), may be deduced from Formulas (4), (6), (11), and 
(13), by substituting 90° — a, for a, and then making the 
proper reductions. 



Collecting the preceding Formulas, we have the following 
table : 

TABLE II. 



(1.) sin 3 a + C03 3 <z 

( 2.) sin 2 a 

( 3.) cos 2 a 

(4.) vcr-sin a 

( 5.) co-vcr-sin a 

(6.) tan a 

(7.) cot a 

(8.) tan a cot a 



1, 




1 — 


cos 2 a. 


1 — 


siu 2 a. 


1 — 


C03 a. 


1 — 


sin a. 


sin 


a 


C03 


• 

a 


COS 


a 


sin 


• 

a 


1. 





( 9.) tan a 

(10.) cot a 

( 11.) sec a 

( 12.) coscc a 

(12.) scc 2 a 

(14. cosec'a 





cot a 




1 




tan a ' 


= 


1 
cos a 


-s 


1 
sin a 


— 


1 + tan'tf 


— 


1 + cot'a 



GO 



ANALYTICAL 



FUNCTIONS OF NEGATIVE AECS. 

62. Let AM"\ estimated from A towards 7>, be 
numerically equal to AM ; then, 
if we denote the arc AM by a, rp/ p 

the arc AM '" will be denoted 
by — a (Art. 48). 

All the functions of AM"\ 
will be the same as those of 
ABM'" \ that is, the functions of 
— a are the same as the func- 
tions of 360° — a. 

From an inspection of the fig- 
ure, we shall discover the following relations, viz. : 

sin (—a) — — sin a ; cos (— a) = cos a ; 
tan (— a) = —tana; cot (— a) = — cot a ; 
sec (— a) — sec a ; cosec (— a) = — cosec a. 




Mt"' 



FUNCTIONS OF ARCS FORMED BY ADDING AN ARC TO, OR SUB- 
TRACTING IT FROM ANT NUMBER OF QUADRANTS. 

63. Let a denote any arc less than 90°. From what 
has preceded, we know that, 

sin (90° — a) = cos a ; cos (90° — a) = sin a. 

tan (90° - a) = cot a; cot (90° — a) = tan a. 

sec (90° — a) = cosec a ; cosec (90° — a) = sec a. 

Now, suppose that J33I' = a, then will A M ' = 90° + a. 
We see from the figure that, 



TRIGONOMETRY. 61 

KM' — sin a, P'M' = cos a, J3T" = tan a, 
AT"' = cot a, OT" = sec a, 02 7 '" = cosec a, 
without reference to their signs. 

By a simple inspection of the figure, observing the rul 
for signs, we deduce the following relations : 

sin (90° -f a) = cos «, cos (90° + a) = - sin a, 

tan (90° + a) = — cotan a, cot (90° + a) = — tan «, 

sec (90° + a) = — cosec a, cosec (90° + a) = sec a. 

Again, suppose 

M'C — AM = a ; then will AM' = 180° — a. 

We see from the figure that, 

P'M'= sin a, OP' = cos a, -4T'" = tan a, 

P7 7 " = cot a, OT" = sec a, OT 1 " = cosec a, 

without reference to their signs : hence, we have, as before, 
the following relations : 

sin (180° - a) = sin «, cos (180° - a) = - cos a, 
tan (180° -a) = - tan « v cot (180° - a) = - cot a, 
sec (180° -a) = - sec a, cosec (180 - a) = cosec a, 

By a similar process, we may discuss the remaining arcs 
in question. Collecting the results, we have the following 

table : 

21 



62 



ANALYTICAL 



TABLE III. 



Arc = 90° + a. 



sin = cos a, cos = — sin a 
tan = 
sec = 



cot a, cot = — tan a, 
cosec a, cosec = sec a. 



Arc 



180 c 



sin 
tan 

sec 



sin a, cos 
tan a, cot 



a. 



= — cos a 



sec a, 



Arc 



cosec 



— cot a, 
cosec a. 



180° + a. 



sin = — sin a, cos 
tan = 
sec = 



— cos a, 



tan a, cot = cot a, 
sec a, cosec == — cosec a. 



Arc = 270° - a. 



sin 
tan 
aec 



cos 



— sin a, 



cos a, 

cot a, cot = tan a, 

cosec a, cosec = — sec a. 



sin 
tan 
sec 



Arc = 270° -+ a. 



cos a, cos = 



cot a, cot = 
cosec a, cosec = 

Arc = 360° - a. 



sin a, 
tan a, 
sec a. 



sin 
tan 
sec 



— sin a, 

— tan a, 
sec a, 



cos 



= cos a, 



cot == — cot a, 
cosec = — cosec a. 



It will be observed that, when the arc is added to, or 
subtracted from, an even number of quadrants, the name of 
the function is the same in both columns ; and when the 
arc is added to, or subtracted from, an odd number of quad- 
rants, the names of the functions in the two columns are 
.contrary: in all cases, the algebraic sign is determined by 
•the rules already given (Art. 58). 

By means of this table, we may find the functions of 
amy arc in terms of the functions of an 'arc less than 90° 

'Thus, 

sin 115° = sin ( 90° -f 25°) = cos 25°, 

sin 284° = sin (270° + 14°) = — cos 14°, 

sin 400° = sin (3G0° -f 40°) = sin 40°, 

tan 210° = tan (180° + 30°) = tan 30°. 



TRIGONOMETRY 




PARTICULAR VALUES OF CERTAIN FUNCTIONS. 

64. Let MAM' be any arc, denoted 
by 2a, M'M its chord, and OA a 
radius drawn perpendicular to M'M: then 
will PM = PM', and AM = AM' 
(B. HI, P. VI.)- But PM is the sine 
of AM, or, PM — sin a : hence. 

sin a = ^M'M ; 

that is, the sine of an are is equal to one half the chord 

of twice the arc. 

Let M'AM = 60° ; then will AM = 30°, and M'M 
will equal the radius, or 1 : hence, we have, 

sin 30° = J ; 

that is, the sine of 30° is equal to half the radius. 

Also, __ 

cos 30° = ^/T^StfZQ* = \V*\ 

hence, 



tan 30° = 



sin 30 c 
cos~30^ 




Agnin, let M'AM = 90° : then will AM = 45°, and 
M'M = J2 (B. V., P. HI.) : hence, we have, 



Also, 



sin 



cos 



45° = \^l\ 

45° = -/I - sin 2 45° = \^\ 



hence, 



sin 45° 

tan 45° = 775 = l - 

cos 45° 



Many other numerical values might be deduced. 



64 



ANALYTICAL 






FORMULAS EXPRESSING- RELATIONS BETWEEN THE CIRCULAR 
FUNCTIONS OF DIFFERENT ARCS. 

65. Let AB and B31 represent two arcs, having the 
common radius 1 ; denote the first by 
b, and the second by a. From 31 
draw 31P, perpendicular to GA, and 
31N perpendicular to GB ; from N 
draw NP' perpendicular to G A, and 
NL parallel to AG. 

Then, by definition, we shall have, 




PP'A 



P31 = sin (a + b), N31 = sin a, and GIT = cos a. 



From the figure, we have, 

PM = PL + LM. 



« • • 



(1.) 



From the right-angled triangle GP'JSF (Art. 37), we have, 

P'N = CR sin b; 



or, since 



P'JST = PL, PL = cos a sin b. 



Sinrje the triangle 31LN is similar to GP'N, the angle 
ZJ£ZV is equal to the angle P'CJST; hence, from the 
right-angled triangle 31L1ST, we have, 

L31 — MN cos b — sin a cos b ; 

Substituting the values of P31, PL, and L3I, in Equa- 
tion ( 1 ), we have, 

sin (a + b) ~ sin « cos H cos a sin 5 ; • ( &.) 

that is, fAe sine of the sum of two arcs, is equal to the 
sine of the first into the cosine of the second, f>lus the co- 
tine of the first into the shie of the second. 



TRIGONOMETRY. 65 

Since the above formula is true for any values of a and 
b, we may substitute - ft, for b ; whence, 

sin (a - b) = sin a cos ( - b) + cos a sin ( - 5) ; 

but (Art. 62), 

cos ( — ft) = cos ft, and, sin ( - 5) = — sin 6 ; 

hence, 

sin (a — ft) = sin a cos b — cos a sin b ; • ( 3.) 

that is, the sine of the difference of two arcs, is equal to 
the sine of the first into the cosine of the second, minus the 
cosine of the first into the sine of the second. 

, If, in Formula (3), we substitute (90° - a), for a, we 

have, 
sin(90°-a-ft) = sin (90° -a) cos b- cos (90° -a) sin ft; • (2.) 

but (Art. 63), 

I sin (90°- a - J) = sin [90°- (a + ft)] = cos (a + 5), 

and, 

sin (90° - «) = cos a, cos (90° - a) = sin a ; 

hence, by substitution in Equation (2), we have, 

cos (a + ft) = cos a cos b — sin a sin b ; • ( ®.) 

\hat is, *Ae coamfi of the sum of two arcs, is equal to the 
ectangle of their cosines, minus the rectangle of their si~ies. 

If, in Formula (®), we substitute - b, for b, we find 

cos (a - b) = cos a cos ( - ft) - sin a sin ( - ft), 
or 

cos (a - ft) = cos a cos ft + sin a sin ft ; • • (2>.) 






M 



ANALYTICAL 



that is, the cosine of the difference of two arcs, is equal 
to the rectangle of their cosines, plus the rectangle of their 
sines. 

If we divide Formula (&) by Formula (®), member by 
member, we have, 

sin (a 4- b) sin a cos b -f cos a sin b 
cos (a + b) cos a cos b — sin a sin b 



Dividing both terms of the second member by cos a cos b, 
recollecting that the sine divided by the cosine is equal to 
the tangent, we find, 

tan a + tan b 



tan (a ■+ b) = : r ; 

v ' 1 — tan a tan b 



• • 



(12.) 



that is, the tangent of the sum of two arcs, is equal to the 
sum of their tangents, divided by 1 minus the rectangle of 
their tangents 

If, in Formula ( SJ ), we substitute — b, for b, recollect- 
ing that tan (— b) = —tan b, we have, 



tan a — tan b 



tan (a — b) = 7 ; 

x 1 + tan a tan b 



• • 



(<<?.) 



that is, the tangent of the difference of two arcs, is equal 
to the difference of their tangents, divided by 1 plus ^ ie 
rectangle of their tangents. 



In like manner, dividing Formula (©) by Formula (&), 
member by member, and reducing, we have, 



cot (a + b) = 



cot a cot b — ". 
cot a 4- cot b 



(®.) 






TRIGONOMETRY. 67 

and thence, by the substitution of -b, for b, 

cot a cot 1) + 1 / s» \ 



• • 



cot (a-b) = -^JTT^T ; 



FUNCTIONS OF DOUBLE ARCS AND HALF ARCS. 

66. If, in Formulas (&), (0), (3), and (Q), we 
make a = b 9 we find, 

sin 2a = 2 sin a cos a ; • • ■ • (<&'.) 

cos 2a = cos 2 a — sin 2 a ;••••( ©'•) 
2 tan a / ro/ \ 

tan2a = r^i^r ; (i2 ° 

COt 2 « — 1 f a\' \ 

cot 2a = — — T K&') 

2 cot a 

/ 

Substituting in ( 0'), f ^r cos 2 a, its value, 1 - sin 2 a ; and 
afterwards for sin 2 a, its value, 1 - cos 2 a, we have, 

cos 2a = 1 — 2 sin 2 a, 
cos 2a = 2 cos 2 a — 1 ; 

whence, by solving these equations, 

/ 1 — cos 2a . , n 

sin a = y ■ g ■ ;••••( 1-) 



cos 



/ 1 -f cos 2a / 9 n 

a = V ' 2 ^ j 



We also have, from the same equations, 

1 — cos 2a = 2 sin 2 a ; ( 3 -) 

1 + cos 2a = 2 cos 2 a. ( 4 -) 



6$ 



ANALYTICAL 



Dividing Equation (&/)> first h J Equation (4), and then 
by Equation ( 3 ), member by member, we have, 



sin 2 a 



1 -J- cos 2a 

sin 2a 

1 — cos 2a 



= tan a 



cot a. 



(5.) 



(6.) 



Substituting %a, for a, in Equations (1), (2), (5), 
and ( 6 ), we have, 



sin \a 



4"- 



cos a 



• • • 



( v.) 



cos -Ja = 




1 + cos a 



( ®".) 



tan ^a 



sin a 



1 + cos a 



(a") 



cot \a = 



sin a 



cos a 



( <3".) 



Taking the reciprocals of both members of the last two 
formulas, we have also, 



n 1 -f cos a 

cot \a = • ; , 

sin a 



and, tan \a = 



1 — cos a 
sin a 



ADDITIONAL FORMULAS. 



67. If Formulas (<&) and (Q) be first added, member 
to member, and then subtracted, and the same operations be 
performed upon ( © ) and ( 5) ), we shall obtain, 



TRIGONOMETRY. 69 

sin (a + b) + sin {a - b) == 2 sin a cos b ; 

sin + b) - sin (a - b) = 2 cos a sin 5 ; 

cos (a + b) + cos (a — Z>) = 2 cos a cos 5 ; 

cos ( a - b) - cos (a -f b) = 2 sin a sin b. 

If in these we make, 

o + b = p, and a — b — q y 

whence, 

a = £ Cfi + ?), 6 = \ {P~2)y 

and then substitute in the above formulas, we obtain, 

sin p + sin q = 2 sin \ {p + ?) cos J {p - q) • ( 5IS.) 

sin -p _ s in £ r= 2 cos -J- (p + ?) sin £ {p — q) • ( &.) 

cos^ + cos? = 2 cos,} {p+ q) cos | Q> - ?) • (S3.) 

cos (7 — cos^ = 2 sin J (p + ?) sin J (p — g) • ( 53.) 

From Formulas (a) and (l&), by division, Ave obtain, 

sin p — si n g _ cosj{p+9) ^ n i(P~Q) _ tanj(j>— g) § ^ j 
ain> + sin <? "" 6in£Cp + 2) cos i(p-q) tsn.$(p+q) 

That is, *Ae sum of the sines of two arcs is to their dif- 
ference, as the tangent of one half the sum of the arcs is 
to the tangent of one half their difference. 



ro 



ANALYTICAL 



Also, in like manner, we obtain, 

sin p + sin q sin j(p+q) cosjjp— g) 
coap + cosq cos i(p+q) cos i(p—q) 



= tMii(p+q) ■ (2.) 



B'mp-smq = sin jjp-q) cos |(j?+g) = tan i (p _^) . (3.) 



sin-J 



sin i 



cos_p -f- cosg 

sin'^ + sin q 
sin (^> + g) 

sin p — sin q 
sin (i?H-g) 

sin {ji—q) 
sin j? — sin ^ sin £1 



COSi(jp + g) COSi(i9-^) 



■(ff+g) cos|(j?-g) _ cos i(p—q) 



(p+q) coai(p+q) cos i{p+ q) 



sin j-(jP— q) cos j(p+g) 
sm i(p+q) cos i(p+q) 

smjjp—q) cos ^(p-q) 
sin i{p—q) cos £Q>+g) 



sin j(p-q) 
sm i(p-\-q) 

cos j(p—q) 
cos i(p+q) 



. (4.) 



(5.) 



. (6.) 



all of which give proportions analogous to that deduced from 
Formula ( 1 ). 

Since the second members of ( 6 ) and ( 4 ) are the same, 

we have, 

sin p — sin q sin (p + q) 



sin (p — q) sinp + sin q 



(*•> 



That is, Z/ie sm<? 0/ the difference of two arcs is to the 
difference of the sines as the sum of the sines to the sine 
of the sum. 

All of the preceding formulas may be made homogeneous 
in terms of jR, R being any radius, as explained in Art. 
30 ; or, we may simply introduce iv, as a factor, into each 
term as many times as may be necessary to render all of 
its terms of the same degree. 



TRIGONOMETRY. 71 

METHOD OF COMPUTING A TABLE OF NATURAL SINES. 

68, Since the length of the semi-circumference of a circle 
wliose radius is 1, is equal to the number 3.1 415 92 Go ... , 
if we divide this number by 10800, the number of minutes 
in 180°, the quotient, .0002908882 . . . , will be the length 
of the arc of one minute ; and since this arc is so small 
that it does not differ materially from its sine or tangent, 
this may be placed in the table as the sine of one minute 

Formula (3) of Table II., gives, 

. cos 1' = y/\ — sin 2 l/ = .9999999577 • • (1.) 

Having thus determined, to a near degree of approxima- 
tion, the sine and cosine of one minute, we take the first 
formula of Art. 07, and put it under the form, 

sin (a + l) = 2 sin a cos b — sin (a — S), 
and make in this, b = 1', and then in succession, 

a = 1', a = 2', a = 3', a = 4', &c., 



in 2' = 2 sin V cos 1' - sin = .00058177G4 . . . 



and obtain, 
sin 

sin 3' = 2 sin 2' cos 1' - sin 1' = .000872GG4G . . . 
sin 4' = 2 sin 3' cos 1' - sin 2' = .0011C3552G . . . 
sin 5' = &0., 

thus obtaining the sine of every number of degrees and 
minutes from 1' to 45°. 



72 ANALYTICAL TRIGONOMETRY. 

The cosines of the corresponding arcs may be computed 
by means of Equation ( 1 ). 

Having found the sines and cosines of arcs less than 45°, 
those of the arcs between 45° and 90°, may be deduced, 
by considering that the sine of an arc is equal to the cosine 
of its complement, and the cosine equal to the sine of the 
complement. Thus, 

sin 50° = sin (90° - 40°) = cos 40°, cos 50° = sin 40°, 

in which the second members are known from the previous 
computations. 

To find the tangent of any arc, divide its sine by its 
cosine. To find the cotangent, take the reciprocal of the 
corresponding tangent. 

As the accuracy of the calculation of the sine of any arc, 
by the above method, depends upon the accuracy of each 
previous calculation, it would be well to verify the work, by 
calculating the sines of the degrees separately (after having 
found the sines of one and two degrees), by the last pro- 
portion of Art. 67. Thus, 

sin 1° : sin 2° — sin 1° : : sin 2° + sin 1° : sin 3° ; 
sin 2° : sin 3° — sin 1° : : sin 3° -f sin 1° : sin 4° ; &c. 



SPHERICAL TRIGONOMETR 



69. Spherical Trigonometry is that branch of Mathe- 
matics which treats of the solution of spherical triangles. 

In every spherical triangle there are six parts : three sides 
and three angles. In general, any three of these parts being 
given, the remaining parts may be found. 



GENERAL PRINCIPLES. 

70. For the purpose of deducing the formulas required 
in the solution of spherical triangles, we shall suppose the 
triangles to be situated on spheres whose radii are equal 
to 1. The formulas thus deduced may be rendered applica- 
ble to triangles lying on any sphere, by making them homo- 
geneous in terms of the radius of that sphere, as explained 
in Art. 30. The only cases considered will be those in 
which each of the sides and angles is less than 1S0°. 

Any angle of a spherical triangle is the same as the die- 
dral angle included by the planes of its sides, and its moa 
sure is equal to that of the angle included between two 
right lines, one in each plane, and both perpendicular to 
their common intersection at the same point (I>. VI., D. 4). 

The radius of the sphere being equal to 1, each side of 
the triangle will measure the angle, at the centre, subtended 
by it. Thus, in the triangle ABC, the angle at A is 



n 



SPHERICAL 



and 



the same as that in eluded between the planes AOO 

A OB ; and the side a is the 

measure of the plane angle BOG, 

being the centre of the 

sphere, and OB the radius, equal 

to 1. 



71. Spherical triangles, like 
plane triangles, are divided into 
two classes, right-angled spherical 

triangles, and oblique-angled spherical triangles. Each class 
will be considered in turn. 

We shall, as before, denote the angles by the capital 
letters A, B, and (7, and the opposite sides by the small 
letters a, 5, and c. 




FORMULAS USED IN SOLVING RIGHT-ANGLED SPHERICAL 

TRIANGLES. 

72. Let CAB be a spherical triangle, right-angled at A, 
and let be the centre of the 
sphere on which it is situated. 
Denote the angles of the triangle 
by the letters A, B, and (7, 
and the opposite sides by the 
letters «, b, and c, recollecting 
that B and C may change 
places, provided that b and c 
change places at the same time. 

Draw OA, OB, and 0(7, each of which will be equal 
to 1. From B, draw BP perpendicular to OA, and 
from P draw PQ perpendicular to OC ; then join the 
points Q and J?, by the line QB. The line QB will be 
perpendicular to 00 (B. VI, P. VI.), and the angle PQB 




TRIGONOMETRY. 75 

will be equal to the inclination of the planes OCB and 
OCA; that is, it will be equal to the angle C. 
We have, from the figure, 

PB — sin c, OP = cos c, QB = sin a, <9<> = cos a. 

From the right-angled triangles OQP and QPB, we 
have, 

OQ =z OP cos ^40(7; or, cos a = cos c cos 5 • (1.) 
PJ5 = @P sin PQB; or, sin c = sin a sin • (2.) 

OP 

If we- multiply both terms of the fraction -Qpt by OQ-> 

we shall have, 

m = m*%> W » cos (7 = tan (90°- a) tan S. (3.) 

OP 

If we multiply both terms of the fraction Typi by PB-> 

we have, 

S = 5? x J§ ; or? sin * = tan c tan (9o °" (7) - (4,) 

If, in (2), we change c and C, into b and P, we 

have, 

sin & = sin'a sin B (5.) 

If, in ( 3 ), we change b and C, into c and P, we 

have, 

cos P = tan (90° — a) tan c • • • • (0. 

If, in (4), we change b, c, and C, into c, &, and 

P, we have, 

sin c = tan b tan (90°— P) • • • • ( 7.) 



76 SPHERICAL 

Multiplying (4) by (7), member by member, we have, 



sm 



b sin c = tan b tan c tan (90°— B) tan (90°— G)> 



Dividing both members by tan b tan c, we have, 
cos b cos c = tan (90° -_B) tan (90°- G) ; 

and substituting for cos b cos c, its value, cos a, taken 
from ( 1 ), we have, 

cos a = tan (90°-_Z?) tan (90°- G) • • (8.) 

Formula (6) may be written under the form, 

_ cos a sin c 

cos B = -i • 

sin a cos c 



Substituting for cos a, its value, cos b cos c, taken from 
(1 ), and reducing, we have, 



cos B = 



cos b sin c 
sin a 



Again, substituting for sin c, its value, sin a sin (7, taken 
from (2), and reducing, we have, 



cos 



B = cos b sin (7 



• • • 



(9.) 



Changing J5, 5, and <7, in (9), into (7, c, and .£, we 
have, 



cos G = cos c sin 7? • •• 



• • 



(10.) 



These ten formulas are sufficient for the solution of any 
right-angled spherical triangle whatever. 



TRIGONOMETRY. 



77 



NAPIER S CIRCULAR PARTS. 




73. The two sides about the right 
angle, the complements of their opposite 
angles, and the complement of the 
hypothenuse, are called Napier's Circular 
Parts. 

If these parts be arranged in their 
order, as shown in the figure, we see that each part is 
adjacent to two of the others, and that it is separated from 
each of two remaining parts by an intervening part. If any 
part be taken as a middle part, those which are adjacent to 
it are called adjacent parts, and tho^e which are separated 
from it, are called opposite parts. Thus, 90° — B, and 
90° — C, are adjacent parts to 90° — a ; and c and b are 
opposite parts ; and so on, for each of the other parts. 



74. Formulas (1), (2), (5), (9), and (10), of 
Art. 12, may be written as follows : 



sin (90°- a) 

sin c 

sin b 

sin (90°-i?) 

sin (90°- C) 



= cos b cos c • ( 1.) 

= cos (90°— a) cos (90°— C) • (2.) 

= cos (90°- a) cos (90°— JO) • (3.) 

= cos b cos (90°— C) .... (4.) 

= cos c cos (90°— JB) .... (5.) 



Comparing these formulas with the figure, we sec that. 

The sine of the middle part is equal to the rectangle of 

the cosines of the opposite parts. 

22 



78 



SPHERICAL 



Formulas (8), (7), (4), (6), and (3), of Art. 12, 
may bo written as follows : 



sin (90°— a) 


— 


tan (90°— .£) (tan 90° ■ 


-(7) 


• (6.) 


sin c 


■ — 


tan b tan (90°— I?) • 


• • 


• (»•) 


sin b 


— 


» 

tan c tan (90°— C) • 


• • 


• (8-) 


sin (90°-i?) 


— 


tan (90°— a) tan c • 


• • 


• (9-) 


sin (90° -C) 


— 


tan (90°— a) tan b • 


• • 


• (io.) 



Comparing these formulas with the figure, we see that, 

The sine of the middle part is equal to the rectangle of 
he tangents of the adjacent parts. 



These two rules are called Napier's rules for Circular 
Parts, and they are sumcient to solve any right-angled 
spherical triangle. 



75. In applying Napier's rules for circular parts, the part 
•sought will be determined by its sine. Now, the same sine 
corresponds to two different arcs, supplements of each other ; 
it is, therefore, necessary to discover such relations between 
■the given and required parts, as will serve to point out 
which of the two arcs is to be taken. 

Two parts of a spherical triangle are said to be of the 
tsame species, when they are both less than 90°, or both 
greater than 90° ; and of different species, when one is less 
and the other greater than 90°. 

From Formulas (9) and (10), Art. 72, we have, 



sin C = 



cos JB 

cos b ' 



and sin B 



cos C 

cos c 



TRIGONOMETRY. 79 

since the angles B and C are both less than 180°, their 
sines must always be positive : hence, cos B must have 
the same sign as cos b, and the cos C must have the 
same sign as cos c. This can only be the case when B 
is of the same species as b, and C of the same species 
as c ; that is, the sides about the right angle are always 
of the same species as their opposite angles. 

From Formula ( 1 ), we see that when a is less than 
90°, or when cos a is positive, the cosines of b and c 
will have the same sign ; that is, b and c will be of the 
same species. "When a is greater than 90°, or when cos a 
is negative, the cosines of b and c will be contrary ; that 
is, b and c will be of different species : hence, ivhen the 
hypothenuse is less than 90°, the two sides about the right 
angle, and consequently the two oblique angles, will be of the 
same species / when the hypothenuse is greater than 90°, 
the tico sides about the right angle, and consequently the two 
oblique angles, will be of different species. 

These two principles enable us to determine the nature 
of the part sought, in every case, except when an oblique 
angle and the opposite side are given, to find the remaining 
parts. In this case, there may be two solutions, one solu- 
tion, or no solution at all. 

Let JBAC be a ri^ht-an- 
gled triangle, in which JB 
and b are given. Prolong 
the sides BA and BC till 
they meet in B'. Take 
B'A' = BA, B'C = BC, and join A' and C by the 
arc of a great circle : then, because the triangles BA C and 
B'A'C have two sides and the included angle of tho one, 
equal to two sides and the included angle of the other, each 
to each, the remaining parts will be equal, each to each ; 





80 SPHERICAL 

that is, A'C = AG, and the angle A' equal to the 
angle A : hence, the two triangles JB A 0, JB'A'C, are 

right-angled ; they have also 
one oblique angle and the op- 
posite side, in each, equal. 

Now, if b differs more from 
90° than J>, there will evident- 
ly be two solutions, the sides 
including the given angle, in the one case, being supplements 
of those which include the given angle, in the other case. 

If I '_J .jj 9 the triangle will be bi-rectangular, and there 
will be but a single solution. 

If b differs less from 90° than J5, the triangle cannot be con- 
structed, that is, there will be no solution. 

SOLUTION OF EIGHT-ANGLED SPHERICAL TRIANGLES. 

76. In a right-angled spherical triangle, the right angle 
is always known. If any two of the other parts are given, 
the remaining parts may be found by Napier's rules for 
circular parts. Six cases may arise. There may be given, 

I. The hypothenuse and one side. 

II. The hypothenuse and one oblique angle. 

TTT, The two sides about the right angle. 

IV. One side and its adjacent angle. 

V. One side and its opposite angle. 

VI. The two oblique angles. 

In any one of these cases, we select that part ^hich is 
either adjacent to, or separated from, each of the other given 
parts, and calling it the middle part, we employ that one of 
Napier's rules which is applicable. Having determined a third 
part, the other two may then be found in a similar manner. 



! 



TRIGONOMETRY. 



81 



It is to be observed, that the formulas employed are to be 
rendered homogeneous, in terms of B, as explained in Art. 30. 
The method of proceeding will be readily understood from a 
few examples. 

EXAMPLES. 



Given a = 105° 17' 29' 



and 
-B, 



1. VjrlYUIl U = 1W J- t *-*J 1 

b = 38° 47' 11", to find c 
and (7. 

Since a > 90°, & and c must be 
of different species, that is, c > 90° ; 
for the same reason, C > 90°. 

OPERATION. 

From Formula (10), Art. 74, we have, 

log cos C = log cot a 4- log tan I — 10; 

log cot a (105° 17' 29") 9.436811 
log tan b ( 3S° 47' 11") 9.903055 




log cos C 



9.34 18G6 



C = 102° 41 ; 33". 



From Formula (2), Art. 74, we have, 

log sin c = log sin a -f log sin (7—10; 



log sin a (105° 17' 29") 9.98434G 

log sin C (102° 41' 33") 9.0892 56 

W sin c 9.973602 



c = 109° 46' 32". 



From Formula (4), we have, 

log cos B = log sin C 4- log cos 5 — 10; 



log sin O (102° 41' 33") 9.9S9256 

log cos b ( 38° 47' 11") 9.891808 

loo- cos B .... 9.881064 



B = 40° 29' 50". 



Ans. c = 109° 46' 32", 5 = 40° 29' 50", C = 102° 41' 33". 



g2 SPHERICAL 

2. Given b = 51° 30', and ^ = 58° 35', to find a, 
c, and (7. 

Because b < B, there are two solutions. 

OPERATION. 

From Formula (7), we have, 

log sin c = log tan b -f log cot I? — 10 ; 

log tan b (51° 30') • 10.099395 
log cot B (58° 35') • 9.785900 

lo"- sin c - - • • 9.885295 .*. c = 50° 09' 51", 

and c = 129° 50' 09". 

From Formula ( 1 ), we have, 

log cos a — log cos b -f- log cos c — 10 ; 

log cos 5 (51° 30') • • 9.794150 
log cos c (50° 09' 51") 9.80658 

log cos a .... 9.600730 .*. a = 66° 29' 54", 

' and a = 113° 30' 06". 

From Formula (10), we have, 

loo- cos G = log tan & 4- log cot a — 10 ; 



log tan b (51° 30') • 10.099395 
log cot a (66° 29' 54") 9.638336 

'loo- cos G .... 9.737731 .*. G = 56° 51' 38", 



an 



d G = 123° 08' 22". 



In a similar maimer, all other cases may be solved. 

3. Given a = 86° 51', and B = 18° 03' 32", to find 
ft, c, and C. 

Ans. b = 18° 01' 50", c = 86° 41' 14", (7 = 88° 58' 25". 



TRIGONOMETRY. 83 

4. Given b = 155° 21' 54", and c = 29° 46' 08", to 
find a, B, and G. 

Ans. a = 142° 09' 13", B = 137° 24' 21", (7 = 54° 01' 16". 

5. Given c = 73° 41' 35", and B = 99° 17' 33", to 
find a, 5, and C. 

Ans. a = 92° 42' 17", 5 = 99° 40' 30", O = 73° 54' 47". 



6. Given 5 = 115° 20', and i? = 91° 01' 47", to find 

o, c, and (7. 

64° 41' 11", fl77° 49' 27", f 177° 35' 36". 

c=\ C = i 

115° 18' 49", [ 2° 10' 33", L 2 ° 24 ' 24 "- 

7. Given I? = 47° 13' 43", and G = 126° 40' 24", to 
find a, J, and c. 

Ans. a = 133° 32' 2G', b == 32° 08' 56", c = 144° 27' 03". 

In certain cases, it may be necessary to find but a single 
part. This may be effected, either by one of the formulas 
given in Art. 74, or by a slight transformation of one of 
them. 

Thus, let a and B be given, to find G. Regarding 
90° _ a, as a middle part, we have, 

cos a = cot B cot G ; 

whence, 

_, cos a 

cot G = — — 7 -> ; 
cot B 

and, by the application of logarithms, 

log cot G = log cos a + (a. c.) log cot B ; 

from which G may be found. In like manner, other cases 
may be treated. 



84 



SPHERICAL 



QUADRANTAL SPHERICAL TRIANGLES. 

77. A Quadrantal Spherical Triangle is one in which 
one side is equal to 90°. To solve such a triangle, we pass 
to its polar triangle, by subtracting each side and each 
ano-le from 180° (B. IX., P. VI.). The resulting polar tri- 
angle will be right-angled, and may be solved by the rules 
already given. The polar triangle of any quadrantal triangle 
being solved, the parts of the given triangle may be found 
by subtracting each part of the polar triangle from 180°. 

EXAMPLE. 

Let A'B'C be a quadrantal 
triangle, in which B'C = 90°, 
B' — 15° 42', and c' = 18° 37'. 

Passing to the polar triangle, 
we have, 

A = 90°, h = 104° 18', and G = 161° 23'. 




Solving this triangle by previous rules, we find, 
a = 1Q° 25' 11", c = 161° 55' 20", B = 94° 31' 21" ; 

hence, the required parts of the given quadrantal triangle are, 



A' - 103° 34' 49", C = 18° 04' 40", V = 85° 28' 39". 



In a similar manner, other quadrantal triangles may be 
solved. 



TRIGONOMETRY. 



85 




FORMULAS USED IN SOLVING OBLIQUE-ANGLED SrHEElCAL TRI- 
ANGLES. 

78. Let ABC represent an oblique-angled spherical tri- 
angle. From either vertex, C, 
draw the arc of a great circle 
CB\ perpendicular to the oppo- 
site side. The two triangles 
ACB' and BCB' will be right- 
angled at B' . 

From the triangle ACB\ we 
have Formula (2), Art. 74, 

sin CB' = sin A sin b. 

From the triangle BCB\ we have, 

sin CB' — sin B sin a. 

Equating these values of sin CB', we have, 

sin A sin b = sin J? sin a ; 

from which results the proportion, 

sin a : sin b : : sin J. : sin B . . . ( 1.) 



In like manner, we may deduce, 

sin a : sin c : : sin A : sin C 
sin b : sin c : : sin B : sin C 



(2.) 
(3.) 



That is, in any spherical triangle, the sines of the sides 
are proportional to the sines of their opposite angles. 

Had the perpendicular fallen on the prolongation of AB^ 
the same relation would have been found. 



86 



SPHERICAL 



79. Let ABC represent any spherical triangle, and 
the centre of the sphere on 
which it is situated. Draw the 
radii OA, OB, and OC\ from 
C draw CJP perpendicular to 
the plane A OB ; from jP, the 
foot of this perpendicular, draw 
BB and BE respectively per- 
pendicular to OA and OB ; join 
CB and CE, these lines will be respectively perpendicular 
to OA and OB (B. VI., P. VI), and the angles CBP 
and CEP will be equal to the angles A and B respec- 
tively. Draw BL and BQ, the one perpendicular, and the 
other parallel to OB. We then have, 

OE = cos a, BG = sin b, OB = cos 5. 




We have from the figure, 



OE = 0B + QB 



(1.) 



In the right-angled triangle OBB, 

OB = OB cos i>(9Z = cos b cos c. 

The right-angled triangle BQB has its sides respectively 
perpendicular to those of OBB ; it is, therefore, similar to 
it, and the angle QBB is equal to c, and we have, 



QB = BB sin QBB = BB sin c 



(2.) 



The right-angled triangle CBB gives, 

BB - CB cos CBB = sin b cos A 5 

substituting this value in ( 2 ), we have, 



$P = sin b sin c cos A ; 



TRIGONOMETRY. 87 

and now substituting these values of OE, OZ, and QP, 
in (1), we have, 

cos a = cos b cosc + sin b sin c cos A • (3.) 

In the same way, we may deduce, 

cos b — cos a cos c + sin a sin c cos J? • • (4.) 
cos c = cos a cos b + sin a sin 5 cos C • • ( 5.) 

That is, «Ae coswie of either side of a spherical triangle is 
equal to the rectangle of the cosines of the other two sides 
plus the rectangle of the sines of these sides into the cosine 
of their included angle. 

80. If we represent the angles of the polar triangle of 
ABC, by A\ B\ and C, and the sides by a\ b' 
and c', we have (B. IX., P. VI.), 

a _ 180 ° - A\ b = 180° - B\ C = 180° - (7', 
"J. = 180° - a\ B = 180° -5', tf = 180° - c'. 

Substituting these values in Equation (3), of the preceding 
article, and recollecting that, 

cos (180°— ^L') = - cos A\ sin (180°- B') = sin B\ &C, 

we have, 

- cos A* = cos B* cos (7' - sin B' sin C" cos a' ; 

or, changing the signs and omitting the primes (since the 
preceding result is true for any triangle), 

cos A = sin B sin G cos a - cos B cos C ( 1.) 



88 



SPHERICAL 



In the same way, we may deduce, 

cos B — sin A sin C cos b — cos A cos C 



cos 



G — sin A sin B cos c — cos ^4. cos B 



(2.) 

(3.) 



That is, the cosine of either angle of a spherical triangle 
is equal to the rectangle of the sines of the other two 
angles into the cosine of their included side, minus the 
rectangle of the cosines of these angles. 

81. From Equation (3), Art. 79, we deduce, 

. cos a — cos b cos c , n x 

cos A = -. — ? — r v. l.) 

sm o sm c 

If we add this equation, member by member, to the num. 
ber 1, and recollect that 1 + cos A, in the first member, 
is equal to 2 cos 2 \A (Art. 66), and reduce, we have, 



sin b sin c 4- cos a — cos b cos c 
2 cos 2 |J. — , 7 _. : 



siD h sin c 

V 

or, Formula ((B), Art. 66, 

cos a — cos (b + c) 



2 cos 2 ^A = 



sin b sin c 



(2.) 



And since, Formula (SI), Art. 67, 

cos a — cos (b + c) = 2 sin l(a + 5 + c) sin -J(6 + c — a), 



Equation ( 2 ) becomes, after dividing both members by 2 

, . sin Ua -f b + c) sin 4(5 + c — a) 

cos 2 *-4 = ■ — t — ' 

1 sin b sm c 



TRIGONOMETRY. 89 

If, in this we make, 
i(a + b + c) = is ; whence, i(b + c - a) = £i - a, 

and extract the square root of both members, we have, 



/ sin is Bin (|s — a) , > 

cos x<a. — Y sin J sm c 

That is, the cosine of one-half of either angle of a spherical 
triangle, is equal to the square root of the sine of one-half 
of the sum of the three sides, into the sine of one-half this 
sum minus the side opposite the angle, divided by the rect- 
angle of the sines of the adjacent sides. 

If we subtract Equation (l), of the preceding article, 
member by member, from the number 1, and recollect that, 

1 _ cos A = 2 sin 2 iA, 
we find, after reduction, 



*n =x/^SS?^ • • • ^ 



sin b sin c 

Dividing the preceding value of sin i A, by cos i A, 

we obtain, 

f^n (js - b) sin (js -j) f g ^ (5<) 
tan^ = y ein^rsln (^ - a) 

82. If the angles and sides of the polar triangle of 
ABC be represented as in Art. 80, we have, 

A = 180° -a\ b = 180° 1 2?', = 180° - (7', 

i8 = 2W - i(A'+B'+ C% is-a = 90°-K*'+ C'-^O. 



90 



SPHERICAL 



Substituting these values in ( 3 ), Art. 81, and reducing 
by tlie aid of the formulas in Table HI., Art. 63, we find, 



sin \a* 



Placing 



r=Goai(A'+B'+C') cos jiB'+G' -A') 



\ 



sin B' sin G 



§{A'+B'+ C) = $S; whence, i(B'+ G'-A') = \8-A f . 



Substituting and omitting the primes, we have, 



sin \a = \A 



— cos j-$ cos (■§■# — A) 



sin B sin (7 



(i.) 



In a similar way, we may deduce from (4), Art. 81. 



cos \a 



-v 



cos (jS-JB) cos (jS- G) 
sin j5 sin G 



(2.) 



and thence, 



tan \a = 



/ — cos|# cos (|£ — A ) 
V cos (i#-i?) cos (iJS- G) 



(3.) 



83. From Equation ( 1 ), Art. 80, we have, 

Q-jTl Jk 

cos A + cos B cos G = sin _Z> sin (7 cos a = sin (7 -^ sin 5 cos & j 



sin a 



(io 



since, from Proportion (1), Art. 78, we have, 



. n sm A . T 

sm B = -. sin o. 

sin a 



Also, from Equation ( 2 ), Art. 80, we have, 



sin j4 

cos i? + cos A cos (7 = sin A sin (7 cos b = sin (7 - sin a cos b. 

sm a 

(2.) 



TRIGONOMETRY. 91 

Adding (1) and (2), and dividing by sin <7, we obtain, 

1 4- cos C sin A . , , v , q > 

(cos A + cos B) -^-^ = ^ sni (a + ft). ( 3.) 

The proportion, sin A : sin B : : sin a : sin ft, 

taken first by composition, and then by division, gives, 

sin A + sin B = ^— (sin a -f sin ft) • • • ( 4.) 

■ sin « 

sin ^- sin B = gpj (sin a - sin 5) - • • (5.) 
Dividing (4) and (5), in succession, by (3), we obtain, 

sin A + sini? ^^L^L = sm_«J-_sin_5 _ > ^ 
35T2 + cosi? 1 + cos 6 f sin {a + ft) 

sin ^ - sin B __sin_C__ _ sin a - s in ft ( ^ 

c^s"^l + cos 5 1 + cos (7 sin (a + ft) 

But, by Formulas (2) and (4), Art. 67, and Formula (SI"), 
Art. 66, Equation (6) becomes, 

, ~ cos $(a— ft) / Q x 

and, by the similar Formulas (3) and (5), of Art. 67, 
Equation (7) becomes, 

, _ sin J(o — ft) / x 

tan^-I?) = cotiCgl^. ■ • W 

These last two formulas give the proportions known as the 
first set of Napier's Analogies. 

cos-Ka + ft) *• cos-K«-ft) • : coi i° : tanK^ + J?). ( 10 «) 
sinK« + ft) ' smi-(a-ft) :: cot^C : Uml(A-B). (11.) 



92 



SPHERICAL 



If in these we substitute the values of a, b, C, A, 
and J3, in terms of the corresponding parts of the polar 
triangle, as expressed in Art. 80, we obtain, 



cos i(A -\-JB) : cos%(A— J3) : : tan^c : 
8m%(A+JB) : sin £( A— B) :: tan-Jc : 
the second set of Napier's Analogies, 



tan-J(a + 5). (12.) 
tan^(a— 5). (13.) 



In applying logarithms to any of the preceding formulas, 
they must be made homogeneous, in terms of R, as ex- 
plained in Art. SO. 



SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. 

84. In the solution of oblique-angled triangles six differ- 
ent cases may arise : viz., there may be given, 

I. Two sides and an angle opposite one of them. 

II. Two angles and a side opposite one of them. 

III. Two sides and their included angle. 

IY. Two angles and their included side. 

V. The three sides. 

VI. The three angles. 

CASE I. 
Given two sides and an angle opposite one of them, 

85. The solution, in this case, is commenced by rinding 
the angle opposite the second given side, for which purpose 
Formula ( 1 ), Art. 78, is employed. 

As this angle is found by means of its sine, and because 
the same sine corresponds to two different arcs, there would 
seem to be two different solutions. To ascertain when there 
are two solutions, when one solution, and when no solution 
at all, it becomes necessary to examine the relations which. 



TRIGONOMETRY. 



93 




may exist between the given parts. Two cases may arise, 
viz., the given angle may be acute, or it may be obtuse. 

We shall consider each case separately (B. IX., P. XIX., 
Gen. Scholium). 

First Case. Let A be 
the given angle, and let a 
and b be the given sides. 
Prolong the arcs AC and 
AB till they meet at A', 
forming the lune AA' ; and 
from (7, draw the arc CB' perpendicular to ABA'. From 
(7, as a pole, and with the arc a, describe the arc of a 
small circle BB. If this circle cuts ABA\ in two points 
between A and A\ there will be two solutions / for if 
C be joined with each point of intersection by the arc of 
a great circle, we shall have two triangles ABC, both of 
which will conform to the conditions of the problem. 

If only one point of in- 
tersection lies between A 
and A\ or if the small 
circle is tangent to ABA\ 
there will be but one solu- 
tion. 

If there is no point of intersection, or if there are points 
of intersection which do not lie between A and A\ there 
will be no solution. 




From Formula ( 2 ), Art. 72, we have, 

sin CB' = sin b sin A, 



from which the perpendicular, which will be less than 90°, 

will be found. Denote its value by p. By inspection of 

the figure, we find the following relations °. 

23 



94: 



SPHERICAL. 




1. When a is greater than p, and at the same time less 
than both b and 180° — b, there will be two solutions. 

2. When a is greater than p, and intermediate in value 
between b and 180° — b; or, when a is equal to p, there 
will be but one solution. 

If a — b, and is also less than 180° — b, one of the points 
of intersection will be at A, and there will be but one 
solution. 

3. When a is greater than p, and at the same time 
greater than both b and 180° — b ; or, when a is 
less than p, there will be no solution. 

Second Case, Adopt the 
same construction as before. 
In this case, the perpendicu- 
lar will be greater than 90°, 
and greater also than any 
other arc CA, CB, CA', 
that can be drawn from C 

to ABA'. By a course of reasoning entirely analogous to 
that in the preceding case, we have the following principles : 

4. When a is less than p, and at the same time 
^greater than both b and 180° — b, there will be two 
solutions. 

5. When a is less than p, and intermediate hi 
■ value between b and 180° — b ; or, ivhen a is equal 
Ho p, there will be but one solution. 

6. When a is less than p, and at the same time 

-less than both b and 180° — b ; or, when a is 

greater than p, there will be no solution. 

•% 

Having found the angle or angles opposite the second 

side, the solution may be completed by means of Napier'a 

Analogies. 



TRIGONOMETRY. 95 



EXAMPLES. 

1. Given a = 43° 27' 36", b = 82° 58' 17", and 
A =3 29° 32' 29", to find B, C, and c. 

We see at a glance, that a > p, since £> cannot 
exceed A ; we see further, that a is less than both b 
and 180° — b ; hence, from the first condition there will be 
two solutions. 

Applying logarithms to Formula (1), Art. 78, we have, 

log sini? = log sin b + log sin ,4 + (a. c.) log sin a - 10 ; 

log sin b • • (82° 58' 17") • • • 9.996724 

log sin A • • (29° 32' 29") - • • 9.692893 

(a. c.) log sin a • • (43° 27' 36") • • • 0.162508 

log sin B 9.852125 

.-. B = 45° 21' 01", and B = 134° 38' 59". 

From the first of Napier's Analogies (10), Art. 83, we find, 

log cot \G — log cos \{a + b) + log tan %(A+B) 

+ (a. c.) log cos i(a — b) — 10. 

Taking the first value of B, we have, 

\(A + B) = 37° 26' 45"; 
also, 

4(a +b) = 63° 12' 56" J and, i{a - b) = 19° 45' 20" 

log cos i(« -f I) • (63° 12 / 56") • 9.653825 

log tan i{A + B) ■ (37° 26' 45") • 9.884130 

(a. c.) log cos i{a - b) • (19° 45' 20") • 0.026344 

logCOtiC 9.564290 

.-. IC = 69° 51' 45", and C sa 139°43'30' f . 



96 SPHERICAL 

The side c may be found by means of Formula ( 12 ), 
Art. 83, or by means of Formula (2), Art. 78. 
Applying logarithms to the proportion, 

sin A : sin G : : sin a : sin c, we hare, 

log sin c = log sin a + log sin G + (a. c.) log sin A — 10 ; 

log sin « ( 43° 27' 36") 9.837492 

log sin G (139° 43' 30") 9.810539 

(a. c.) log sin^L ( 29° 32' 29") 0.30710 7 

log sin c 9.955138 .'. c — 115° 35' 48". 

We take the greater value of c, because the angle G, 
being greater than the angle i?, requires that the side c 
should be greater than the side b. By using the second 
value of B, we may find, in a similar manner, 

G = 32° 20' 28", and c = 48° 16' 18". 

2. Given a = 97° 35', b ±= 27° 08' 22", and 
A = 40° 51' 18", to find B, G, and c. 

J.ws. B - 17° 31' 09", G = 144° 48' 10", c = 119° 08' 25". 

3. Given a = 115° 20' 10", 5 = 57° 30' 06", and 
A = 126° 37' 30", to find i?, (7, and c. 

Ans. B — 48° 29' 48", G — 61° 40' 16", c = 82° 34' 04". 

CASE n. 
Given two angles and a side opposite one of them. 

86. The solution, in this case, is commenced by finding 
the side opposite the second given angle, by means of For- 
ula (1), Art. 78. The solution is completed as in Case I. 



m 



TRIGONOMETRY. 97 

Since the second side is found by means of its sine, there 
may be two solutions. To investigate this case, we pass to 
the polar triangle, by substituting for each part its supple- 
ment. In this triangle, there will be given two sides and 
an angle opposite one ; it may therefore be discussed as in 
the preceding case. When the polar triangle has two solu- 
tions, one solution, or no solution, the given triangle will, 
in like manner, have two solutions, one solution, or no solu- 
tion. 

The conditions may be written out from those of the pre- 
ceding case, by simply changing angles into sides, and the 
reverse ; and greater into less, and the reverse. 

Let the given parts be A, B, 
and a, and let p be an arc 
computed from the equation, 




sm p 



= sin a sin B. 



There will be two cases : a may be greater than 99° ; 
or, a may be less than 90°. 

In the first case, 

1. When A is less than p, and at the same time 
greater than both B and 180° - B, there will be two 
* solutions. 

2. When A is less than p, and intermediate in 
value between B and 180° — B ; or, when A is equal 
to p, there will be but one solution. 

3. When A is less than p, and at the same time 
less than both B and 180° - B / or, when A U 
greater than p, there will be no solution. 



98 " SPHERICAL 

In the second case, 

4. When A is greater than p, and at the same 
less than both B and 180° - B, there will be two solu 
tions. 

5. When A is greater than p, and intermediate in 
value between B and 180° - B ; or, when A is equal 
to p, there will be but one solution, 

6. When A is greater than p, and at the same 
time greater than both B and 180° - B ; or, when A 
is less than p, there will be no solution. 

EXAMPLES. 

1. Given A = 95° 16', B = 80° 42' 10", and 
a = 57° 38', to find «, b, and C. 

Computing p, from the formula, 

log sin p = log sin B + log sin a — 10 ; 
we have, p = 56° 27' 52". 

The smaller value of p is taken, because a is less 

than 90°. 

Because A > p, and intermediate between 80° 42' 10" 
and 99° 17' 50", there will, from the fifth condition, be but 
a single solution. 

Applying logarithms to Proportion ( 1 ), Art. 78, we have, 

log sin b == log sin B -f log sin a + (a. c.) log sin A — 10 ; 

log sin 77 (80° 42' 10") 9.994257 

log sin a (57° 38') 9.926671 

(a.c.) log sin A (95° 16') 0.001837 

log sin b .... 9.922765 .'. b = 56° 49' 57" 



TRIGONOMETRY. 99 

We take the smaller value of b, for the reason that A, 
being greater than B, requires that a should be greater 

than b. 

Applying logarithms to Proportion (12), Art. 83, we have,. 

log tan Jc = log cos i(A + B) + log tan \{a + b) 

+ (a. c.) log cos i(A - B) - 10 ; 
we have, 

i(A + B) = 87° 59' 05", \{a + b) = 57° 13' 58", 
and, VA - B) = 7° 16' 55". 



log cos i{A + B) ■ (87° 59' 05") . 8.546124 

ligtantfa + ft) - (57° 13' 58") ■ 10.191352 

(a. c.) log cos i(A - B) • (7° 16' 55") . 0.003517 

log tan \c 8.740993 

... \c = 3° 09' 09", and c = 6° 18' 18". 

Applying logarithms to the proportion, 

sin a : sin c : : sin A : sin (7, 
we have, 

log sin G = log sin c + log sin J. + (a. c.) log sin a — 10 ; 

log sin c (6° 18' 18") . 9.040685 

log sin A (95° 16') • • 9.998163 

(a. c.) log sin a (57° 38') • • 0.073329 

log sin G 9.112177 .*. (7= 7° 20' 21". 

The smaller value of C is taken, for the same reason 
as before. 

2. Given A = 50° 12', B = 58° 08', and a = 62°42', 
to find 5, c, and G. 

79° 12' 10", [119° 03' 20", f 130° 54' 28", 

1 = ^ 100° 47' 50", ° ~ J 152° 14' 18", " \ 156° 15' 06". 



100 SPHERICAL 

case in. 

Given two sides and their included angle. 

87. The remaining angles are found by means of Kapier^ 
Analogies, and the remaining side, as in the preceding cases. 

EXAMPLES. 

1. Given a = 62° 38', b = 10° 13' 19", and 

C = 150° 24' 12", to find c, A, and B. 

Applying logarithms to Proportions (10) and (11), 
Art. 83, we have, 

log tan ^(A + JB) = log cos i(a - b) 4- log cot \G 

+ (a. c.) log cos \{a -f b) — 10 ; 

log tan \{A-JB) = log sin %(a - b) + log cot J<7 

+ (a. c.) log sin |(a + b) — 10 ; 
we have, 

i(a-b) = 26° 12' 20", |(7 = 75° 12' 06", 

and, i(« + 5) = 36° 25' 39". 

log cos i(a - 6) ■ (26° 12' 20") - 9.952897 

logcoti<7 • • • (75° 12' 06") • 9.421901 

(a. c.) log cos i(a + b) • (36° 25' 39") ■ 0.094415 

log tan i(A + B) 9.469213 

.-. i{A + JB) = 16° 24' 51". 

log sin i{a - b) • (26° 12' 20") - 9.645022 

logcot£<7 • • • (75° 12' 06") • 9.421901 

(a. C.) log Sin i(a +b) • (36° 25' 39") • 0-226356 

log timl{A-J3) 9.293279 

,\ i(A -JS) = IP 06' 53". 



TRIGONOMETRY. 101 

The greater angle is equal to the half sum plus the half 
difference, and the less is equal to the half sum minus the 
half difference. Hence, we have, 

A = 27° 31' 44", and B = 5° 17' 58". 

Applying logarithms to the Proportion (13), Art. 83, we 

have, 

log tan \c = log sin \{A + B) + log tan \{a - b) 

+ (a. c.) log sin \(A - B) '.- 10 ; 

log sin i(A + B) ■ (16° 24' 51") ■ 9.451139 

log tan i(a - 5) • (26° 12' 20") • 9.692125 

(a. c.) log sin i(A - B) • (11° 06' 53") - 0.714952 

log tan \c 9.858216 

.-. \c = 35° 48' 33", and c = 71° 37' 06". 

2. Given a = 68° 46' 02", b — 37° 10', and 
C = 39° 23' 23", to find c, A, and J>. 

4rc*. Jl = 120° 59' 47", B = 33° 45' 03", c = 43° 37' 38". 

3. Given a = 84° 14' 29", b = 44° 13' 45", and 
C = 36° 45' 28", to find J. and B. 

Arts. A = 130° 05' 22", B = 32° 26' 06". 



CASE IV. 
Given two angles and their included side. 

88. The solution of this case is entirely analogous to that 
of Case III. 

Applying logarithms to Proportions (12) and (13 j, Art. 
83, and to Proportion (11), Art. 83, we have, 



102 SPHERICAL 

log tan i(a + b) = log cos \{A — B) + log tan \c 

4- (a. c.) log cos ^(J. + i?) — 10 ; 

log tan \{a — b) = log sin \{A — B) -f log tan -Jc 
I + (a. c.) log sin %(A -f B) — 10 ; 

log cot -J (7 = log sin -J(a + #) + log tan J( A — J5) 

+ (a. c.) log sin \{a — b) — 10 ; 

The application of these formulas are sufficient for the 
solution of all cases. 

EXAMPLES. 

1. Given A = 81° 38' 20", B = 70° 09' 38", and 
c = 59° 16' 22", to find G, a, and b. 

Ans. G = 64° 46' 24", a = 70° 04' 17", 5 = 63° 21' 27". 

2. Given ^i = 34° 15' 03", -S = 42° 15' 13", and 
c = 76° 35' 36", to find G, a, and b. 

Ans. G = 121° 36' 12", a = 40° 0' 10", b = 50° 10' 30". 



CASE V. 
Given the three sides, to find the remaining parts. 

89. The angles may be found by means of Formula ( 3 ), 
Art. 81 ; or, one angle being found by that formula, the other 
two may be found by means of Napier's Analogies. 

EXAMPLES. 

1. Given a == 74° 23', b = 35° 46' 14", and c = 100° 39', 
to find A, B, and G. 



TRIGONOMETRY. 103 

Applying logarithms to Formula (3), Art. 81, we have, 

log cos \A = 10 + i P°g sm i s + l°g sm (i s ~~ a ) 

+ (a. c.) log sin b + (a. c.) log sin c — 20] ; 
or, 

log cos \A = i [log sin Js + log sin (is — a) 

+ (a. c.) log sin b + (a. c.) log sin c], 
we have, 

js = 105° 24' 07", and \s — a = 31° 01' 07". 

log sin Js • • • (105° 24' 07") • 9.984116 

log sin {\s -a) - ( 31° 01' 07") • 9.712074 

(a. c.) log sin 5 .... ( 35° 46' 14") - 0.233185 

(a. c.) log sin c (100° 39') 0.007546 

2)10.936921 
log cos \A 9.968460 

.-. \A — 21° 34' 23", and A = 43° 08' 46". 

Using the same formula as before, and substituting B for 
A, b for a, and a for b, and recollecting that 
is — b — 69° 37' 53", we have, 

log sin is • • • (105° 24' 07") . 9.984116 

log sin (is - b) • ( 69° 37' 53") - 9.971958 

(a. c.) log sin a • • • • . (74° 23') • • 0.016336 

(a. c.) log sin c (100° 39') • • 0.007546 

2)19.979956 
log COS \B 9.089978 

.-. iZ? = 12° 15' 43", and B = 24° 31' 26'. 

Using the same formula, substituting C for A f c for a, 
and a for c, recollecting that is — c s= 4° 45' 07", we 
have, 



104: SPHERICAL TRIGONOMETRY. 



log sin -Js 



log sin (^s 
(a. c.) log sin a ■ 
(a. c.) log sin b « 



. • (105° 24' 07") 

- c) • (4° 45' 07") 

. • • (74° 23') • 

. . (35° 46' 14") 



9.984116 

8.918250 

. 0.016336 

• 9.233185 
2)19.151887 



log cos J G 



• • ¥ 



±C 



9.575943 

67° 52' 25", and G = 135° 44' 50". 



2. Given a — 56° 40', h = 83° 13', and c = 114° 30'. 
J.W5. A = 48° 31' 18", B = 62° 55' 44", (7 = 125° 18' 56". 



CASE VI. 
^TAe ^Aree angles being given, to find the sides. 

90. The solution in this case is entirely analogous to the 
preceding one. 

Applying logarithms to Formula ( 2 ), Art. 82, we have, 

cos QS - B) + log cos {iS-C) 

":>g sin G J. 



log cos \a 



4 [log 



" ~ — v * 

+ (a. c.) log 



'5 <^° V2^ ^ , 

■ sin i> + (a. c.) log sii 



In the same manner as before, we change the letters, to 
suit each case. 



EXAMPLES. 



1. Given A = 48° 30', B = 125° 20', and G == 62° 54'. 
4»S. a = 56° 39' 30", b = 114° 29' 58", c = 83° 12' 06" 

2. Given ^L = 109° 55' 42", B = 116° 38' 33", and 
C = 120° 43' 37", to find «, J, and c. 



^m. a = 98° 21' 40", 5 = 109° 50' 22", c = 115° 13' 28". 



MENSURATION. 



91. Mensuration is that branch of Mathematics which 
treats of the measurement of Geometrical Magnitudes. 

92. The measurement of a quantity is the operation of 
finding how many times it contains another quantity of the 
same kind, taken as a standard. This standard is called the 
unit of measure. 

93. The unit of measure for surfaces is a square, one 
of whose sides is the linear unit. The unit of measure for 
volumes is a cube, one of whose edges is the linear unit. 

If the linear unit is one foot, the superficial unit is one 
square foot, and the unit of volume is one cubic foot. If 
the linear unit is one yard, the superficial unit is one square 
yard, and the unit of volume is one cubic yard. 

94. In Mensuration, the term product of two lines, is 
used to denote the product obtained by multiplying the 
number of linear units in one line by the number of linear 
units in the other. The term product of three Hues, is used 
to denote the continued product of the number of linear 
nnits in each of the three lines. 

Thus, when we say that the area of a parallelogram is 
equal to the product of its base and altitude, we mean that 
the number of superficial units in the parallelogram is equal 
to the number of linear units in the base, multiplied by the 
number of linear units in the altitude. In like manner, the 



106 MENSURATION 

number of units of volume, in a rectangular parallelopipedon, 
is equal to the number of superficial units in its base multi- 
plied by the number of linear units in its altitude, and 
so on. 

MENSURATION OF PLANE EIGUEES. 
To find the area of a parallelogram, 

95. From the principle demonstrated in Book IV., 
Prop. V., we have the following 

RULE. 

Multiply the base by the altitude; the product will be 
the area required. 



EXAMPLES. 



1. Find the area of a parallelogram, whose base is 12.25, 
and whose altitude is 8.5. ■ Ans. 104.125. 

2. What is the area of a square, whose side is 204.3 
f eet ? Ans. 41738.49 sq. ft. 

3. How many square yards are there in a rectangle, 
whose base is 66.3 feet, and altitude 33.3 feet ? 

A?i$. 245.31 sq. yd. 

4. What is the area of a rectangular board, whose 
length is 12 \ feet, and breadth 9 inches ? 9} sq. ft. 

5. What is the number of square yards in a parallelo- 
gram, whose base is 37 feet, and altitude 5 feet 3 inches? 

Ans. 21 T y 

To find the area of a plane triangle. 
96. First Case. When the base and altitude are given. 



OF SURFACES. 



107 



From the principle demonstrated in Book X , Prop. VI., 
we may write the following 



RULE. 

Multiply the base by half the altitude ; the product wiU 
be the area required. 

examples. 

1. Find the area of a triangle, whose base is G25, and 
altitude 520 feet. -*»■ 16250 ° ** ft ' 

2. Find the area of a triangle, in square yards, whose 
.base is 40, and altitude 30 feet. ^ns. 66*. 

3. Find the area of a triangle, in square yards, whose 
base is 49, and altitude 25^ feet. Am. 68.7361. 

Second Case. When two sides and their included angle 
are given. 

Let ABC represent a plane tri- 
angle, in which the side AB — c, 
BC = a, and the angle B, are 
given. From A draw AB perpen- 
dicular to BC ; this will be the 
altitude of the triangle. From For- 
mula ( 1 ), Art. 37, Plane Trigonometry, we have, 

AB = c sin B. 

Denoting the area of the triangle by ft and *VVV™S tbe 
rule last given, we have, 

ac sin B 




Q = 



2 



or. 



2 = ac sin B. 



sin B 



Substituting for sin S, ^f (Trig., Art. 30), and applying 

logarithms, we have, 

log (2Q) = log a + log c + log sin B — 10 ; 



108 



MENSURATION" 



hence, we may write the following 

pule. 

Add together the logarithms of the two sides and the 
logarithmic sine of their included angle; from this sum 
subtract 10 ; the remainder will be the logarithm of double 
the area of the triangle. Find, from the table, the number 
answering to this logarithm, and divide it by 2 ; the quotient 
will be the required area. 

EXAMPLES. 

1. What is the area of a triangle, in which two sides 
a and b, are respectively equal to 125.81, and 57.65, and 
whose included angle C, is 57° 25' ? 

Ans. 2Q — 6111.4, and Q — 3055.7 Arts. 

2. What is the area of a triangle, whose sides are 30 
and 40, and their included angle 2S° 57' ? Ans. 290.427. 

3. What is the number of square yards in a triangle, of 
which the sides are 25 feet and 21.25 feet, and their included 
angle 45° ? Ans. 20.8694. 



LEMMA. 

To find half an angle, when the three sides of a plane tri- 
angle are given. 

97. Let ABC be a plane tri- 
angle, the angles and sides being de- 
noted as in the figure. 

We have (B. XV., P. XII., XIII.), 




a 



2 = b 2 + c 2 =f 2c . AD. 



(1.) 



When the angle A is acute, we have (Art. 37), 

AD = b cos A ; when obtuse, AD' = b cos CAD'. 



OF SURFACES. 109 

But as CAB' is the supplement of the obtuse angle A, 
cos CAB' = - cos A, and AB' — — b cos A. 

Either of these values, being substituted for AB, in ( 1 ), 

gives, 

a 2 = b 2 + c 2 — 25c cos J. ; 

whence, 

fr* + c 2 - a 2 , v 

COsJ " = 263 W 

If we acid 1 to both members, and recollect that 
1 + cos^i = 2 cosH^- (Art. 6G), Equation (4), Ave have, 

, , ( 26c + b 2 + c 2 - a 2 

2 cos 2 ^! = r^- 

1 2bc 

_ (^ + c) 2 - a 2 _ (b i- c + a) (& + g - a) . 

~~ 26c "~ 26c ' 

or, 



cos 



A (b±^±a)_ (b + c - _g) # _ b ^ (8>) 

46c 



If we put 6 + c + a = 5, Ave have, 

^±±^ = h and, L+*H2 = J, _ a , 

Substituting in (3), and extracting the square root, 



cos £.4 = V 6c ' • • ' • ( 4 

the plus sign, only, being used, since, \A < 90° ; hence, 

The cosine of half of cither angle of a plane triangle, 
is equal to the square root of half the sum of the three 
sides, into half that sum minus the side opposite the angle, 
divided by the rectangle of the adjacent sides. 

By applying logarithms, we have, log cos -M = 

\ [log \s + log (U - a) + (a. c.) log b + (a. c.) log c]. • ( &.) 

24 



110 MENSURATION 

If we subtract both members of Equation ( 2 ), from 1, 
and recollect that 1 - cos A = 2 sin 2 \A (Art. 37), we have, 

, ■ \ , 2bc — b 2 — c 2 + a 2 

2 BUI' U --= ■ We 

«2 _ (5 _ c) 2 (a + b — c) (a — b + e) . . 

26c 26c v 

Placing, as before, a + 6 + c = *, we have, 

— — = As — c, and, r = %s — o. 

2 2 

Substituting in (5), and reducing, we have, 



hence, V &> ■ 

^TAc sirce c/ Aa?/" cm angle of a plane triangle, is equal 
to the square root of half the stem of the three sides, minus 
,one of the adjacent sides, into the half sum minus the 
other adjacent side, divided by the rectangle of the adjacent 
sides. 

Applying logarithms, we have, 

log sin iA. = 'i [log (is -b) + log (§s - c) 

+ (a. c.) log b + (a. c.) log c]. (Q.) 



Third Case. To find the area of a triangle, when the 
"hrce sides are given. 

Let ABC represent a triangle 
whose sides a, b, and c are given. 
From the principle demonstrated in 
the last case, we have, 

Q = \bc sin A, 




OF SURFACES. Ill 

But, from Formula (&')', Tri S-> Art - 66 > we hav8 > 

sin A — 2 sin \A cos \A ; 

whence, . 

Q — be sin ^4. cos $-4... 

Substituting for sin j2 and cos ^ their values, taken 
from Lemma, and reducing, we have, 

Q = Vi* (is-«) (i s -^) (^- c )^ 
hence, we may write the following 

RULE. 

Find half the sum of the three sides, and from it subtract 
each side separately. Find the continued product of the half 
sum and the three remainders, and extract its square root ; tU 
result will be the area required. 

It is generally more convenient to employ logarithms ; for 
this purpose, applying logarithms to the last equation, we have, 

log Q = ipog i» + log (i« - a) -r log fa - b) 4- log (**-"«)] 

hence, we have the following 

EULE. 

Find the half sum and the three remainders as before, then 
find the half sum of their logarithms ; the number correspond- 
ing to the resulting logarithm icill be the area required. 

EXAMPLES. 

1. Find the area of a triangle, whose sides are 20, 30, 

and 40. 

We have, \s = 45, U-a = 25, U-b = 15, Ja-C = G. 

By the first rule, 

Q = y r 45^ 25 x 15 x 5 = 290.4737 Am. 



112 



MENSURATION 



By the second rule, 



log is 


• • • 


• ( 45 ) • 


. • • 1.653213 


log (is- 


— a) 


• (25) • 


. • • 1.397940 


log {is ■ 


-b) ■ 


• (15) 


. . • 1.176091 


log (is ■ 


-c) ■ 


• (5) • 


. • • 0.698970 
2 )4. 926214 


lo<? C 


) . . . 


• • • 


. • • 2.463107 



.-. Q = 290-4737 Ans. 

2. How many square yards are there in a triangle, whose 
sides are 30, 40, and 50 feet ? Ans. 66|. 



To find the area of a trapezoid. 

98. From the principle demonstrated in Book IV., Prop. 
VII., we may write the following 

RULE. 

Find half the sum of the parallel sides, and multiply it 
by the altitude ; the product will be the area required. 

E XAMPLE S. 

* 

1. In a trapezoid the parallel sides are 750 and 1225, 
and the perpendicular distance between them is 1540 ; what 
is the area ? Ans. 1520750. 

2. How many square feet are contained in a plank, whose 
length is 12 feet 6 inches, the breadth at the greater end 15 
inches, and at the less end 11 inches? A?is. 13^. 

3. How many square yards are there in a trapezoid, 
whose parallel sides are 240 feet, 320 feet, and altitude 66 
feet ? Ans - 2053 i *<l- T& 

To find the area of any quadrilateral. 
99, From what precedes, we deduce the following 



OF SURFACES. 



113 




EULE. 

Join the vertices of two opposite angles by a diagonal; 
from each of the other vertices let fall perpendiculars upon 
this diagonal; multiply the diagonal by half of the sum 
of the perpendiculars, and the product will be the area re- 
quired. 

E X AMP LE S. 

1. What is the area of the quad- 
rilateral ABCD, the diagonal AG 
bem" 42, and the perpendiculars Dg, 
Bb, equal to IS and 16 feet ? 

Ans. VI 4 sq. ft. 

2. How many square yards of paving are there in the 
quadrilateral, whose diagonal is 65 feet, and the two perpen- 
diculars let fall on it 2S and 33^ feet ? Ans. 222 T V 

To find the area of any polygon. 
100. From what precedes, we have the following 

EULE. 

Draw diagonals dividing the proposed polygon into tra- 
pezoids and triangles : then find the areas of these figures 
separately, and add them together for the area of the whole 
polygon. 

EXAMPLE. 

1. Let it be required to de- 
termine the area of the polygon 
ABODE, having five sides. 

Let us suppose that we have mea- 
sured the diagonals and perpendicu- 
lars, and found AC = 36.21, EG = 39.11, Bb = 4 
Dd - V.2G, Aa = 4.18 : required the area. Ans. 29G.1292. 




114 



MENSURATION 



To find the area of a regular polygon. 

101. Let AB, denoted by 5, re- 
present one side of a regular polygon, 
whose centre is C. Draw CA and 
CB, and from C draw CD perpen- 
dicular to AB. Then will CD he the 
apothem, and Ave shall have AD — BD. 

Denote the number of sides of the polygon by n ; then 




will the angle A CB, at the centre, be equal to 



360 c 



n 



(B. V., Page 138, D. 2), and the angle A CD, which is half 

1 80° 
of ACB, will be equal to 



n 



In the right-angled triangle ADC, we shall have, For- 
mula (3), Art. 37, Trig., 

CD = is tan CAD. 

But CAD, being the complement of A CD, we have, 

tan CAD = cot A CD ; 

180° 



hence, 



CD = *5 cot 



n 



a formula by means of which the apothem may be computed. 
But the area is equal to the perimeter multiplied by half 
the apothem (Book V., Prop. VIII.) : hence the following 

RULE 

Find the apothem, by the preceding formula / multiply 
the perimeter by half the apothem / the product icill be t/ie 
area required. 

EXAMPLES. 

1. What is the area of a regular hexagon, each of whose 
sides is 20 ? We have, 

CD : 10 : : cot 30° ; or, log CD = log 10 + log cot 30° — 10 

1.000000 



lo£ U 



o 2' 



log cot 



180° 



n 

log CD 



(10 ) 
(30°) 



10.2385G1 
1.2385C1 



CD 



17.3205. 



OF SURFACES. 



115 



The perimeter is equal to 120 : hence, denoting the area by §, 

120 X 17.3205 



Q = 



= 1039.23 Ans. 



2, What is the area of an octagon, one of whose side?. 
ig 20 ? Ans - 1931.36886. 

The areas of some of the most important of the regular 
polygons have been computed by the preceding method, on 
the supposition that each side is equal to 1, and the results 
are given in the following 









TABLE. 






1 


NAME-!. 


shje.-. 


AREAS. 


NAMES. 


SIDES. 


AREAS. 




Triangle, ' . 


. 3 


, . 0.4330127 


Octagon, . 


. 8 , 


. 4.S2S4271 




Square, 


. 4 


. 1.0000000 


Xonagon, . 


. 9 . 


. 6.1818242 




Pentagon, . 


. 5 


. 1.7204774 


Decagon, . 


. 10 


. . 7.6942088 




Hexagon . 


. G 


. 2.59807G2 


Undecagon, 


. 11 


. 9.3656399 




Heptagon . 


. 7 


. 3.633912-i 


Dodecagon, 


. 12 


. 11.1961524 



The areas of similar polygons are to each other as the 
squares of their homologous sides (Book IV., Prop. XXVLL). 

Denoting the area of a regular polygon whose side is 
*j by (?> an(1 tnat of a simnar polygon whose side is 
1, by 1\ the tabular area, we have, 

Q : T : : s 2 : l 2 ; .'. Q = Ts*; 



hence, the following rule. 

Multiply the corresponding tabular area by the square of ^ 
the (jiven side ; the product will be the area required. 

EXAMPLES. 

1. What is the area of a regular hexagon, each of whose 

sides is 20 ? 

We have, T = 2.59S07G2, and s 2 = 400 : hence, 

Q = 2.59807G2 x 400 = 1039.23048 Ans. 



116 



MENSURATION 



2. Find the area of a pentagon, whose side is 25. 

Ans. 1075.298375, 

3. Find the area of a decagon, whose side is 20. 

Ans. 3077.68352. 

To find the circumference of a circle, when the diameter is 

given. 

102. From the principle demonstrated in Book V., Prop, 
XVI., we may write the following 

EULE. 

Multiply the given diameter by 3.1416 ; the product loiU 
be the circumference required. 

EXAMPLES. 

1. What is the circumference of a circle, whose diameter 
is 25 ? Ans. 78.54. 

2. If the diameter of the earth is 7921 miles, what is 
the circumference? - Ans. 24884.6136. 

To find the diameter of a circle, when the circumference is 

given. 

103. From the preceding case, we may write the following 

R TJ L E . 

Divide the given circumference by 3.1416 ; the quotient 
will be the diameter required. 



E X AMPLES. 

1. What is the diameter of a circle, whose circumference 
is 11652.1944 ? Am. 3709. 

2. What is the diameter of a circle, whose circumference 
is 0850 ? Ans. 2180.41 



OF SURFACES. 117 

To find the length of an arc containing any number of 

degrees. 
104. The length of an arc of 1°, in a circle whose 
diameter is 1, is equal to the circumference, or 3.1 41 G 
divided by 3G0 ; that is, it is equal to O.00872GG : hence, 
the length of an arc of n degrees, will he, n X 0.0087266. 
To find the length of an arc containing n degrees, when 
the diameter is d, we employ the principle demonstrated in 
Book V., Prop. XIII., C. 2 : hence, we may write the following 

EULE. 

Multiply the number of degrees in the are by .0087266, 
and the product by the diameter of the circle ; the resxdt 
will be the length required. 

examples. 

1. What is the length of an arc of 30 degrees, the 

diameter being 18 feet? ' Ans. 4.7123G4 ft. 

•2. What is the length of an arc of 12° 10', or 12}°, the 

diameter being 20 feet ? Ans. 2.123472 ft. 

To find the area of a circle. 

105. From the principle demonstrated in Book V., Prop. 
XV., we may write the following 

RULE. 

Multiply the square of the radius by 3.1416 ; the pro- 
duct will be the area required. 

EXAMPLES. 

1. Find the area of a circle, whose diameter is 10, and 
circumference 31.416. Ans. 78.54. 

2. How many square yards in a circle whose diameter 
is 3J feet? Ans - 1.069016. 

3. What is the area of a circle whose circumference is 
12 feet ? Ans ' H-4M5. 



118 



MENSURATION 



To find the area of a circular sector. 

106, From the principle demonstrated in Book V., Prop. 
XIV., C. 1 and 2, we may write the following 

EULE. 

I. Multiply half the arc by the radius ; or, 
II. Find the area of the whole circle, by the last rule; 
then write the proportion, as 360 is to the member of degrees 
in the sector, so is the area of the circle to the area of tlie 
sector. 

E XAMPLE S. 

1. Find the area of a circular sector, whose arc contains 
18°, the diameter of the circle being 3 feet. 0.35343 sq. ft. 

2. Find the area of a sector, whose arc is 20 feet, the 
radius being 10. Ans. 100. 

3. Required the area of a sector, whose arc is 147° 29', 
and radius 25 feet. Ans. 804.3986 sq. ft. 

To find the area of a circular segment. 

107. Let AB represent the chord 
corresponding to the two segments 
ACB and AFB. Draw AE and 
BE. The segment ACB is equal to 
the sector EACB, minus the triangle 
AEB. The segment AFB is equal 
to the sector EAFB, plus the tri- 
angle AEB. Hence, we have the fol- 
lowing 

EULE. 

Find the area of the corresponding sector, and also of 
the triangle formed by the chord of the segment and the 
two extreme radii of the sector ; subtract the latter from the 
former when the segment is less than a semicircle, and talce 
their sum when the segment is greater than a semicircle; 
the result will be the area required. 




OF SURFACES. 119 



EXAJIP LES. 



1. Find the area of a segment, whose chord is 12 and 
the radius 10. 

Solving the triangle AEB, we find the angle AEB ia 
equal to 73° 44', the area of the sector EACB equal to 
64.35, and the area of the triangle AEB equal to 48 ; 
hence, the segment ACB is equal to 1G.35 Ans. 

2. Find the area \)f a segment, whose height is IS, the 
diameter of the circle being 50. Ans. 636.4834. 

3. Required the area of a segment, whose chord is 16, 
the diameter being 20. Ans. 44.764. 



To find the area of a circular ring contained between the 
circumferences of two concentric circles. 

108. Let B and r denote the radii of the two circles, 
B being greater than r. The area of the outer circle is 
B 2 X 3.1416, and that of the inner circle is r 2 X 3.1416 ; 
hence, the area of the ring is equal to (B z — r 2 ) X 3.1416. 
Hence, the following 

EULE. 



Find the difference of the squares of the radii of the 
two circles, and multiply it by 3.1416 ; the product icill be 
the area required. 

EXAMPLES. 

1. The diameters of two concentric circles being 10 and 
6, required the area of the ring contained between their 
circumferences. Ans. 50.2056. 

2. "What is the area of the ring, when the diameters of 
the circles are 10 and 20 ? Ans. 235.62.. 



120 MENSURATION 

MENSURATION OF BROKEN AND CURVED SURFACES. 
To find the area of the entire surface of a right prism. 

109. From the principle demonstrated in Book TIL, Prop, 
L, we may write the following 

RULE. 

Multiply the perimeter of the base by the altitude, the pro- 
duct will be the area of the convex surface ; to this add the 
areas of the two bases ; the result to ill be the area required. 

EXAMPLES. 

1. Find the surface of a cube, the length of each side 
being 20 feet. Ans. 2400 sq. ft. 

2. Find the whole surface of a triangular prism, whose 
base is an equilateral triangle, having each of its sides equal 
to 18 inches, and altitude 20 feet. Ans. 91.949 sq. ft. 

To find the area of the entire surface of a right pyramid. 

110. From the principle demonstrated in Book VIL, Prop. 
IV., we may write the following 

RULE. 

Multiply the perimeter of the base by half the sla?it 

height; the product will be the area of the convex surface; 

to this add the area, of the base; the result will be the area 
required. 

EXAMPLES. 

1. Find the convex surface of a right triangular pyramid, 
the slant height being 20 feet, and each side of the bass 
3 feet. Ans. 90 sq. ft 

2. What is the entire surface of a right pyramid, whose 
slant height is 15 feet, and the base a pentagon, of which 
each side is 25 feet ? Ans. 2012.798 sq. ft. 



OF SURFACES. 121 

To find the area of the convex surface of a frustum of a 

right 'pyramid. 

111. From the principle demonstrated in Book XII., Prop. 
IV., C, we may write the following 

E TJ L E . 

Multiply the half sum of the perimeters of the two bases 
by the slant height; the product will be the area required 

EXAMPLES. 

1. How many square feet are there in the convex sur- 
face of the frustum of a square pyramid, whose slant height 
is 10 feet, each side of the lower base 3 feet 4 inches, and 
each side of the upper base 2 feet 2 inches? Ans. 110 sq.ft. 

2. What is the convex surface of the frustum of a 
heptagonal pyramid, whose slant height is 55 feet, each side 
of the lower base 8 feet, and each side of the upper base 
4 f ee t ? Ans. 2310 sq. ft. 

112. Since a cylinder may be regarded as a prism whose 
base has an infinite number of sides, and a cone as a pyra- 
mid whose base has an infinite number of sides, the rules 
just given, may be applied to find the areas of the surfaces 
of right cylinders, cones, and frustums of cones, by simply 
changing the term perimeter, to circumference. 

EXAMPLES. 

1. What is the convex surface of a cylinder, the diameter 
of whose base is 20, and whose altitude 50? Ans, 3141.6 

2. What is the entire surface of a cylinder, the altitude 
bcinn- 20, and diameter of the base 2 feet? 131.9472 sq.ft. 

3. Required the convex surface of a cone, whose slant 

height is 50 feet, and the diameter of its base 8.1 feet. 

Am. G07.50 sq. ft. 



122 MENSURATION 

4. Required the entire surface of a cone, whose slant 
height is 36, and the diameter of its base 18 feet. 

Am. 1272.348 sq. ft. 

5. Find the convex surface of the frustum of a cone, the 
slant height of the frustum being 12| feet, and the circum- 
ferences of the bases 8.4 feet and 6 feet. Am. 90 sq. ft. 

6. Find the entire surface of the frustum of a cone, the 
slant height being 16 feet, and the radii of the bases 3 feet, 
and 2 feet. ^ns. 292.1688 sq. ft. 

To find the area of the surface of a sphere. 

113. From the principle demonstrated in Book VIII, 
Prop. X., C. 1, we may write the following 

RULE. 

Find the area of one of its great circles, and multiply 
it by 4 ; the product will be the area required. 

EXAMPLES. 

1. What is the area of the surface of a sphere, whose 
radius is 16? ' ^ ns - 3216.9984. 

2. What is the area of the surface of a sphere, whose 
radius is 27.25 ^ s - 9331.3374 

To find the area of a zone. 

114. From the principle demonstrated in Book VIII., 
Prop. X., C. 2, we may write the following 

RULE. 

Find the circumference of a great circle of the sphere, 
and midtiply it by the altitude of the zone ; the product 
xoill be the area required. 



OF SURFACES. 123 

EX AMP LES. 

1. The diameter of a sphere being 42 inches, what is 
the area of the surface of a zone whose altitude is 9 inches. 

Ans. II 67.5248 sq. in. 

2. If the diameter of a sphere is 12 J feet, what will be 
the surface of a zone whose altitude is 2 feet ? 78.54 sq. ft. 

To find the area of a spherical polygon. 

115. From the principle demonstrated in Book IX., Prop. 
XIX., we may write the following 

RULE. 

From the sum of the angles of the polygon, subtract 1S0° 
taken as many times as the polygon has sides, less two, 
and divide the remainder by 90° ; the quotient will be the 
spherical excess. Find the area of a great circle of the 
sphere, and divide it by 2 ; the quotient will be the area 
of a tri-rectangular triangle. Multiply the area of the tri- 
rectangular triangle by the spherical excess, and the product 
will be the area required. 

This rule applies to the spherical triangle, as well as to 
any other spherical polygon. 

EXAMPLES. 

1. Required the area of a triangle described on a sphere, 
whose diameter is 30 feet, the angles being 140°, 92°, and 
330^ Ans. 471.24 sq. ft 

2. What is the area of a polygon of seven sides, de 
seribed on a sphere whose diameter is 17 feet, the sum of 
the angles being 10S0° ? Ans. 2 ' 20 - 98 

3. What is the area of a regular polygon of eight sides, 
described on a sphere whose diameter is 30 yards, each an- 
gle of the polygon being 140° ? Ans. 157.08 sq. yds. 



124 MENSURATION" 

MENSURATION OF VOLUMES. 
To find the volume of a prism. 

118. From the principle demonstrated in Book VTL, 
Prop. XIV., we may write the following 

EULE. 

Multiply the area of the base by the altitude ; the pro- 
duct will be the volume required. 

EXAMPLES. 

1. What is the volume of a cube, whose side is 24 inches ? 

Arts. 13824 cu. in. 

2. ' How many cubic feet in a block of marble, of which 
the length is 3 feet 2 inches, breadth 2 feet 8 inches, and 
height or thickness 2 feet 6 inches ? Arts. 21£ cu. ft. 

3. Required the volume of a triangular prism, whose 
height is 10 feet, and the three sides of its triangular base 
3, 4, and 5 feet. (Ans. 60. 

To find the volume of a pyramid. 

117. From the principle demonstrated in Book VII., Prop. 
XVII., we may write the following 

EULE. 

Multiply the area of the base by one-third of the alti- 
tude ; the product will be the volume required. 

EXAMPLES. 

1. Required the volume of a square pyramid, each side 
of its base being 30, and the altitude 25. Ans. 7500. 

2. Find the volume of a triangular pyramid, whose alti- 
tude is 30, and each side of the base 3 feet. 38.9711 cu. ft. 



OF VOLUMES. 125 

3. What is the volume of a pentagonal pyramid, its alti- 
tude being 12 feet, and each side of its base 2 feet. 

Ans. 27.5270 cu. ft. 

4. What is the volume of an hexagonal pyramid, whose 

altitude is G.4 feet, and each side of its base 6 inches ° 

Ans. 1.38564 cu. ft. 

To find the volume of a frustum of a pyramid. 

118. From the principle demonstrated in Book VII., Prop., 
XYHL, C, we may write the following 

EULE. 

Find the sum of the upper base, the lower base, and a 
mean proportional between them ; multiply the result by one- 
third of the altitude ; the product wiU be the volume required. 

EXAMPLES. 

1. Find the number of cubic feet in a piece of timber, 
whose bases are squares, each side of the lower base being 
15 inches, and eacli side of the upper base 6 inches, the 
altitude being 24 feet. ^ns. 19.5. 

2. Required the volume of a pentagonal frustum, whose 
altitude is 5 feet, each side of the lower base 18 inches, and 
each side of the upper base G inches. Ans. 9.31925 cu. ft. 

119. Since cylinders and cones are limiting cases of prisms 
and pyramids, the three preceding rules are equally applicable 
to them. 

EXAMPLES. 

1. Required the volume of a cylinder whose altitude is 

12 feet, and the diameter of its base 15 feet. 

Ans. 2120.58 CU. ft. 

2. Required the volume of a cylinder whose altitude is 
20 feet, and the circumference of whose base is 5 feet 

a . , Ans. 48.144 cu. ft. 

6 inches. 

25 



126 MENSURATION 

3. Required the volume of a cone whose altitude 18 
27 feet, and the diameter of the base 10 feet. 

Ans. 706.86 cu. ft. 

4. Required the volume of a cone whose altitude is 
10 \ feet, and the circumference of its base 9 feet. 

Ans. 22.56 cu. ft. 

5. Find the volume of the frustum of a cone, the altitude 
beino- 18, the diameter of the lower base 8, and that of the 
upper base 4. Ans. 527.7888. 

6. What is the volume of the frustum of a cone, the 
altitude being 25, the circumference of the lower base 20, 
and that of the upper base 10 ? Ans. 464.216. 

7. If a cask, which is composed of two equal conic frus- 
tums joined together at their larger bases, have its bung dia- 
meter 28 inches, the head diameter 20 inches, and the length 
40 inches, how many gallons of wine will it contain, there 
toeing 231 cubic inches in a gallon? Ans. 79.0613. 

To find the volume of a sphere. 

120. From the principle demonstrated in Book VIII., 
IProp. XIV., we may write the following 

RULE. 

Cube the diameter of the sphere, and multiply the result 
Iby itf, that is, by 0.5236 ; the product will be the volume 
'.required. 

EXAMPLES. 

1. What is the volume of a sphere, whose diameter is 
12 y Ans. 904.7808 

2. What is the volume of the earth, if the mean diam 

eter be taken equal to 7918.7 miles. 

Ans. 259992792083 cu. miles. 



OF VOLUMES, 



127 




H 




To find the volume of a wedge. 

121. A Wedge is a volume bound- 
ed by a rectangle ABCB, called the 
back, two trapezoids ABHG, DCHG, 
called faces, and two triangles ADG, 
GBH, called ends. The line Gil, in 
which the faces meet, is called the edge. 
The two faces are equally inclined to 
the back, and so also are the two ends. 

There are three cases: 1st, When the length of the edge is 
equal to the length of the back; 2d, When it is less; and 3d, 
When it is greater. 

In the first case, the wedge is a right prism, whose base is 
the triangle ADG, and altitude GIT or AB : hence, its volume 
is equal to ADG multiplied by AB. 

In the second case, through II, 
the middle point of the edge, pass 
a plane II CB perpendicular to the 
back and intersecting it in the line 
BC parallel to AB. This plane 
will divide the wedge into two 
parts, one of which is represented 
by the figure. 

Through G, draw the plane G-NM parallel to IICB, and it 
will divide the part of the wedge represented by the figure into 
the right triangular prism GNM — JB, and the quadrangular pyr- 
amid A&NM— G. Draw GP perpendicular to NM\ it will 
also be perpendicular to the back of the wedge (B. VI., P. 
XVII.), and hence, will be equal to the altitude of the wedge. 

Denote AB by I, the breadth AD by h, the edge Gil by 
I, the altitude by h, and the volume by V\ then, 

AJf=B-l, MB = Gil = I, and area JVGJl=loh: then 

Prism = \bhl\ Pyramid = b{I - t)^h = l s hh(I - I), and 
V= mi + \bh(Z -T) = VM + \hhL - \Ul = lJdt(l+2L). 

We can find a similar expression for the remaining part of the 
wedge, and by adding, the factor within the parenthesis becomes 
the entire length of the edge plus twice the length of the back. 




128 



MENSURATION 



In the third case, I is greater than Z, and denotes the 
altitude of the prism; the volume of each part is equal to 
the difference of the prism and pyramid, and is of the same 
form as before. Hence, the following 

Rule.— Add twice the length of the back to the length q ; 
he edge; multiply the sum by the breadth of the back, and 
that result by one-sixth of the altitude ; the final product will 
be the volume required. 

EXAMPLES. 

1. If the back of a wedge is 40 by 20 feet, the edge 
35 feet, and the altitude 10 feet, what is the volume? 

Ans. 3833.33 cu.ft. 

2. What is the volume of a wedge, whose back is 18 feet 
by 9, edge 20 feet, and altitude 6 feet? 504 cu.fi. 

To find the volume of a prismoid. 

122. A Prismoid is a frustum of a wedge. 

Let L and B denote the 
length and breadth of the lower 
base, I and b the length and 
breadth of the upper base, M and 
m the length and breadth of the 
section equidistant from the bases, 
and h the altitude of the prismoid. 

Through the edges L and l\ 
let a plane be passed, and it will 

divide the prismoid into two wedges, having for bases, the 
bases of the prismoid, and for edges the lines L and V. 

The volume of the prismoid, denoted by I 7 ", will be 
equal to the sum of the volumes of the two wedges ; hence, 



r 




M 




or. 



V = i£h(l + 2i) + }bh(Z + 21) ; 

V = \h{2BL + M +JBI + bZ) ; 



OF VOLUMES. 129 

which may be written under the form, 

V = }h [(BZ + bl + Bl + bZ) + J5Z + bl]. (&.) 

Because the auxiliary section is midway between the bases, 

we have, 

231 = Z+l, and 2m = JB + b ; 

hence 

4Mm = {Z+l) {B + b) = BZ + bl + BZ \- bl 

Substituting in (&), we have, 

V — lh(BL + bl + 4:31m). 

But BL is the area of the lower base, or lower section, 
bl is the area of the upper base, or upper section, and Mm 
is the area of the middle section ; hence, the following 

R U L E . 

To find the volume of a prismoid, find the sum of the 
areas of the extreme sections and four times the middle sec- 
tion ; multiply the result by one-sixth of the distance between 
the extreme sections ; the result will be the volume required. 

This rule is used in computing volumes of earth-work in 
railroad cutting and embankment, and is of very extensive 
application. It may be shown that the same rule holds for 
every one of the volumes heretofore discussed in this work. 
Thus, in a pyramid, we may regard the base as one extreme 
* section, and the vertex (whose area is 0), as the other 
extreme ; their sum is equal to the area of the base. The 
area of a section midway between between them is equal to 
one-fourth of the base : hence, four times the middle section 
is equal to the base. Multiplying the sum of these by one- 
sixth of the altitude, gives the same result as that already 
found. The application of the rule to the case of cylinders, 
frustums of cones, spheres, etc., is left as an exercise for the 
student. 



130 MENSURATION 



EXAMPLE S. 

1. One of the bases of a rectangular prismoid is 25 feet 
by 20, the other 15 feet by 10, and the altitude 12 feet 
required the volume. Ans. 3700 cu. ft. 

2. What is the volume of a stick of hewn timber, 
whose ends are 30 inches by 27, and 24 inches by 18, its 
length, being 24 feet? Ans. 102 cu.-ft. 



MENSURATION OF REGULAR POLTEDRONS. 

123. A Regular Polyedeo:n" is a polyedron bounded bv 
equal regular polygons. 

The polyedral angles of any regular polyedron are all 
equal. 

jL 

124. There are five regular poiyedrons (Book VII., 

Page 208). 

To find the dledral angle between the faces of a regidar 

polyedron. 

125. Let the vertex of any polyedral angle be taken as 
the centre of a sphere whose radius is 1 : then will this 
sphere, by its intersections with the faces of the polyedral 
angle, determine a regular spherical polygon whose sides will 
be equal to the plane angles that bound the polyedral angle, 
and whose angles are equal to the diedral angles between 
the faces. 

It only remains to deduce a formula for finding one 
angle of a regular spherical polygon, when the sides are 
given. 



OF POLYEDRONS. 



131 



Let ABODE represent a regular spherical polygon, and 
let P be the pole of a small circle passing through its verti- 
ces. Suppose P to be connected 
with each of the vertices by arcs of 
great circles ; there will thus be 
formed as many equal isosceles tri- 
angles as the polygon has sides, the 
vertical angle in each being equal 
to 360° divided by the number of 
sides. Through JP draw BQ per- 
pendicular to AB : then will A Q 
be equal to BQ. If we denote the number of sides by w, 




360' 



180 c 



or 



Che angle AP Q will be equal to ^ , n 

In the right-angled spherical triangle APQ, we know the 
base AQ, and the vertical angle APQ ; hence, by Napier's 
rules for circular parts, we have, 

s ; n (oo° _ APQ) = cos (90° - PA Q) cosAQ; 

or, bv reduction, denoting the side AB by s, and the an- 

gle PAB, by A, 

1S0° 



cos 



n 



— sin \ A cos \s ; 



cos 



180° 



whence, 



sin \A = 



n 



cos \s 



EXAMPLES, 



In the Tetracdron, 
180° 



n 



— G0°, and \s = 



In the Ilexaedron, 



180 



n 



- — 60°, and 



30° 



Is 

2* 



45 



••^o 



A - V0° 31' 42". 



A = 90°. 



132 MENSURATION 

In the Octaedron, 

— = 45°, and is = 30° .\ A = 109° 28' 18". 
n 

In the Dodecaedron, 

= 60°, and J* = 54° .-. A = 116° 33' 54". 



n 

In the Icosaedron 
180° 



n 



= 36°, and is = 30° .'. ^. = 138° 11' 23". 



To j^rac? tfAe volume of a regular polyedron. 

126. If planes be passed through, the centre of the poly- 
edron and each of the edges, they will divide the polyedron 
into as many equal right pyramids as the polyedron has faces. 
The common vertex of these pyramids will be at the centre 
of the polyedron, their bases will be the faces of the poly- 
edron, and their lateral faces will bisect the diedral angles 
of the polyedron. The volume of each pyramid will be equal 
to its base into one-third of its altitude, and this multiplied 
by the number of faces, will be the volume of the polyedron. 

It only remains to deduce a formula for finding the dis- 
tance from the centre to one nice of the polyedron. 

Conceive a perpendicular to be drawn from the centre of 
the polyedron to one face ; the foot of this perpendicular 
will be the centre of the face. From the foot of this per- 
pendicular, draw a perpendicular to either side of the free 
in which it lies, and connect the point thus determined with 
the centre of the polyedron. There will thus be formed a 
right-angled triangle, whose base is the apothem of the face, 
whose angle at the base is half the diedral angle of the 
polyedron, and whose altitude is the required altitude of the 
pyramid, or in other words, the radius of the inscribed 
sphere. 



OF POLYEDRONS. 



133 



Denoting the perpendicular by P, the base by 5, and 
the diedral angle by A, we have Formula (3), Art. 37, 

Trig., 

P = b tan \A ; 

hat b is the apothem of one face ; if, therefore, we denote 
the number of sides in that face by n, and the length of 
each side by s, we shall have (Art. 101, Mens,), 

180° 



b = is cot 



n 



whence, by substitution, 



P = 



180 ° 1 A 

|s cot ■ tan f A ; 



n 



hence, the volume may be computed. The volumes of all 
the regular polyedrons have been computed on the supposi- 
tion that their edges are each equal to 1, and the results 
are given in the following 



NAMES. 

Tetraedron, . 
Ilexaedron, . 
Octaedron, 
Dodecaedron, 
Icosaedron, , 



TABLE. 

NO. OP FACES. VOLUMES. 

4 0.1178513 

6 1.0000000 

8 0.4714045 

12 7.GG31189 

20 .... . 2.181G950 



From the principles demonstrated in Book VII., we may 
write the following 



B ULE. 



To find the volume of any regular potyedron, multiply 
tJie cube of its edge by the corresponding tabular volume; 
tJie product will be the volume required. 



134 MENSURATION. 

EXAMPLES. 

1. What is the volume of a tetraedron, whose edge is 15 ? 

Ans. 391,15. 

2. "What is the volume of a hexaedrou, whose edge is 12 ? 

Ans. 1728. 

3. What is the volume of a octaedron, whose edge is 20 ? 

Ans. 3771.236. 

4. What is the volume of a dodecaedron, whose edge 
is 25 ? . Ans. 119736.2328. 

5. What is the volume of an icosaedron, whose edge 
is 20 ? Ans. 17453.56. 



A TABLE 



OF 



LOGARITHMS OF NUMBERS 



FROM 1 TO 10,000. 



N. 



Log. 



3 
4 

5 



9 

10 

ii 

12 

i3 

U 
i5 

16 

iB 
'9 

20 

21 
22 
23 
2 4 
25 



o-oocooo 

0-477I2I 
o- 602060 
0-698970 

- 8 1 5 1 
o- 845098 
0-901090 
0-954243 

I -000000 

1 -04i3q3 
1 .079181 
1 • 1 1 1943 
1 • 1 161 28 

1 -176091 

I -204120 

i«23o4 io 

1 -2" 
1-2" 

1 • 3uio3o 



N. 



I • 3222 10 

I \7 S 
[728 
! $021 I 
1 -397940, 



26 

27 
28 
29 

3o 
3i 

32 

33 
34 
35 

36 

3 

3 

39 
40 

41 
42 
43 
44 
45 

46 

47 
48 

49 
5o 



Lost. 



414973 
43s 364 

447 ' 53 
462398 

477121 

491362 
i-5o5i5o 
1 • 5 1 '• 3 1 4 
•53 1 479 
.544068 

■5563o3 
•568202 
•579784 
.591065 
1 .602060 

1-612784 
1 .62 h ig 

1-6! 

1-643453 

1-6532 1 3 



,662758 

672098 
.681241 
• 690 1 96 
■ 6989 



N. 



5i 

52 

53 
54 
55 

56 

n 

5 9 

60 

61 
62 

63 
64 
65 

66 

6 l 
68 

69 

70 

7i 
72 
73 
74 

_7l 



Loss 



707070 
716003 
724276 
732394 
74o363 

748188 
755875 
763428 
770852 
7781D1 

78533o 
792392 

799341 
■806181 
■812913 

•819544 
.826075 
•8325o9 
■838840 

•845098 

-85i258 
-857 J33 
.863323 

.869232 
■873061 



N. 



76 

77 
78 

79 
80 

81 
82 
83 

84 

85 

86 

*7 
88 
89 
90 

9i 
92 

93 
94 
9 5 

96 

97 

98 

99 
100 



Log. 



1. 880814 

886491 
892095 
897627 
903090 

908485 

■ 9 i38i4 
919078 

•924279 
.929419 

.934498 
•939519 
•944483 
•949390 
.954243 

•959041 
.963788 
.968483 
.973128 

•977724 

1 -9^2271 

I ' 5773 
1-99I220 

I .995635 

■/. .QOOOOO 



Remark. la the following table, in the nine right band 
columns of each page, where the first or leading figures 
chan-e from 9's to O's, points or dots are introduced in- 
stead of the O's, to catch the eye, and to indicate that from 
thence the two figures of the Logarithm to he taken from 
the second column, stand in the next line below. 



A TABLE OF LOGARITHMS FROM 1 TO 10,000. 



N. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


D. 


IOO 


000000 


0434 


0868 


i3oi 


1734 


2166 


2598 


3029 


3461 


38 9 i 


43a 


IOI 


432i 


4721 


5i8i 


5609 


6o38 


6466 


6894 


7321 


7748 


8174 


428 


102 


8600 


9026 
3259 


945i 


9876 


•3oo 


•724 


1 147 


1570 


i 99 3 


24i5 


424 


io3 


012837 


368o 


4100 


452i 


4940 


536o 


5 779 


6197 


6616 


419 


104 


7o33 


743i 


7868 


8284 


8700 


91 16 


9532 


?947 


•36i 


• 77 5 


416 


io5 


021 189 


i6o3 


2016 


2428 


2841 


3252 


366/ 


4075 


4486 


48g6 


412 


106 


53o6 


5710 


6i25 


6533 


6942 


735o 


7737 


8164' 


8D7 1 


8g 7 8 


408 


107 


9 3 84 


9789 


•195 


°6oo 


1004 


1408 


1812 


2216 


2619 


3021 


404 


108 o33424 


3826 


4227 


4628 


5o2 9 


543o 


583o 


623o 


6629 


7028 


400 


109 


7426 


7825 


8223 


8620 


9017 


9414 


9811 


•207 


•602 


•998 


3g6 


no 


041393 


1787 


2182 


25-76 


2969 


3362 


3755 


4U8 


4540 


4g32 


3 9 3 


III 


5323 


5714 


6io5 


6495 


688d 


7275 


7664 


8o53 


8442 


8S3o 


38g 


112 


0218 

053078 


9606 


9993 


•3 80 


•766 


1 1 53 


1 53 8 


1924 


2309 


2694 


386 


Ii3 


3463 


3846 


423o 


46i3 


4996 


5378 


5760 


6142 


6524 


382 


114 


6905 


7286 


7666 


8046 


8426 


88o5 


gi85 


9 563 


9942 


•320 


37? 


u5 


060698 


1075 


1452 


1829 


2206 


2582 


2g58 


3333 


3709 


4o83 


3 7 6 


116 


4458 


4832 


52o6 


558o 


5953 


6326 


6699 


7071 


7443 


78i5 


372 


117 


8186 


855 7 


8928 


9298 


9668 


••38 


•407 


•776 


1 145 


i5i4 


36g 


118 


071882 


225o 


2617 


2g85 


3352 


3718 


4o85 


445 1 


4816 


5 182 


366 


119 


5547 


5912 


6276 


6640 


7004 


7368 


773i 


8094 


8457 


8819 


363 


120 


079181 


9D43 


9904 


•266 


•626 


•987 


1 347 


1707 


2067 


2426 


36o 


121 


082785 


3i44 


35o3 


386i 


4219 


4576 


4934 


5291 


5647 


6004 


357 


122 


636o 


6716 


7071 


7426 


7781 


8i36 


8490 


8845 


9198 


o552 


355 


123 


9go5 


•258 


•611 


• 9 63 


i3i5 


1667 


2018 


2370 


2721 


3o 7 i 


35i 


124 


093422 


3772 


4122 


4471 


4820 


5 1 69 


55i8 


5866 


62i5 


6562 


34g 


125 


6910 


72-57 


7604 


7 9 5i 


8298 


8644 


8990 


9335 


9681 


••26 


346 


126 


100371 


0715 


1059 


i4o3 


H47 


2091 


2434 


2777 


3ii9 


3462 


343 


127 


38o4 


4146 


448-; 


4828 


5i6 9 


55io 


585 1 


6191 


653 1 


6871 


34o 


128 


7210 


7 54g 


7S88 


8227 


8565 


8 9 o3 


9241 


9 5 79 


9916 


•253 


338 


129 


1 10590 


0926 


1263 


1 599 


1934 


2270 


26o5 


2940 


32^5 


0609 


335 


i3o 


1 1 3g43 


4277 


461 1 


4944 


5278 


56n 


5 9 43 


6276 


6608 


6940 


333 


i3i 


7271 


76o3 


7Q34 


8265 


85 9 5 


8926 


9256 


9 586 


99 i5 


•245 


33o 


132 


120574 


0903 


I 23 1 


i56o 


1888 


2216 


2544 


2871 


3198 


3525 


328 


i33 


3852 


4178 


45o4 


483o 


5 1 56 


5481 


58o6 


6i3i 


6456 


6781 


325 


i34 


7io5 


7429 
06 5 D 


77 53 


8076 


83 99 


8722 


9045 


9 368 


9690 


••12 


323 


i35 


i3o334 


0977 


1298 


1619 


i 9 3 9 


2260 


258o 


2900 


3219 


321 


i36 


353 9 


3858 


4177 


4496 


4814 


5i33 


545i 


5769 


6086 


64o3 


3i8 


i3 7 


6721 


7o3 7 


7354 


7671 


79^7 


83 o3 


8618 


8 9 3i 


9249 


9 564 


3i5 


i38 


9879 


•194 


•5o8 


•822 


u36 


i45o 


1763 


2076 


2389 


2702 


3i4 


i3 9 


i43oi5 


3327 


363 9 


395i 


4263 


4574 


4885 


5196 


55o7 


58i8 


3u 


•140 


146128 


6438 


6748 


7 o58 


736 7 


7676 


79 85 


8294 


86o3 


891 1 


309 


U' 


9219 


9D27 


9 835 


•142 


•449 


•i56 


io63 


1370 


1676 


19^2 


307 


142 


152288 


25 9 4 


2900 


32o5 


35io 


38i5 


4120 


4424 


4728 


5o32 


3o5 


143 


5336 


5640 


5g43 


6246 


6549 


6852 


ii 54 


7457 


7759 


S061 


3o3 


144 


8362 


8664 


8 9 65 


9266 


9 56 7 


9868 


•168 


•469 


•7^9 


1068 


3oi 


U5 


161 368 


1667 


1967 


2266 


2 564 


2863 


3i6i 


346o 


3758 


4o55 


2 99 


146 


4353 


46 5o 


4947 


5244 


5541 


5838 


6i34 


643o 


6726 


7022 


297 


147 


7 3i 7 


7613 


7908 
0848 


82o3 


8497 


8792 


9086 


g38o 


9674 


9 g68 


295 


148 


170262 


o555 


1141 


U34 


1726 


2019 


23l I 


26o3 


2Sg5 


2g3 


149 


3i86 


3478 


3769 


4060 


435i 


4641 


4g3 2 


5222 


55i2 


58o2 


£■91 


i5o 


1 7609 1 


638i 


6670 


6o5g 


7248 


7536 


7825 


3n3 


8401 


86^9 


289 


1 5 1 


8977 
181844 


9264 


9552 


9 83 9 


•126 


•4i3 


•699 


• 9 85 


1272 


1 558 


287 


i5a 


2129 


24i5 


2700 


2q85 


3270 


3555 


3S3 9 


4i23 


4407 


285 


1 53 


4691 


4975 


5259 


5542 


5825 


6108 


6391 


6674 


6g56 


7 2 3g 


283 


1 5 4 


7521 


7 8o3 


8084 


8366 


8647 


8928 


9209 


9490 


9771 


••5 1 


281 


i55 


190332 


0612 


0892 


1171 


1 45 1 


n3o 


2010 


2289 


2567 


2846 


279 


156 


3 1 25 


34o3 


368i 


3g5g 


4237 


45 1 4 


4-92 


5069 


5346 


5623 


278 


1 5 7 


58o 9 


6176 


6453 


6729 


7oo5 


7281 


75d6 


7832 


8107 


8382 


276 


1 58 


86^7 


8g32 


9206 


9481 


9755 


••29 


•3o3 


•5 77 


•85o 


1 1 24 


274 


159 


201397 


1670 


1943 


2216 


2488 


2761 


3o33 


33o5 


3577 


3848 


272 
1). 


N. 





1 


2 


3 


4 


5 


6 


7 


8 


9 



A TABLE OF LOGARITHMS FROM 1 TO 10,000. 



3 



N. 



ibo 
161 

162 
i63 
164 
165 
166 
167 
168 
169 

170 
171 
172 
l 7 3 

175 

176 

xi i 
178 

179 

180 
181 
182 

1 83 
184 
1 85 
186 



189 

190 
191 
192 
193 
194 
ig5 
196 
197 
198 
199 

200 
201 
202 
2o3 
204 
2o5 
206 
207 
208 
209 

210 

21 1 
212 

2l3 

214 

21 5 

216 

21 
21 
219 



204120 
6826 

90 '5 

212188 
4844 
7484 

22010S 
2716; 
53o 9 ' 
78871 

23o449 
2990 
5328 
8046 

240349 
3o38 
55i3 

7973 
25o42o 
2853 

2552,73 
7679 

260071 
245i 
4818 
7172 
931 3 

271842 
4i58 
6462 

278754 

2 8io33 

33oi 

5 Tit 

7802 

290033 

2256 

4466 

6665 
8853 



4391 
7006 
97 83 
2454 
5109 

77^7 
0370 
2976 
5568 
8i44 

0704 
325o 
5781 
8297 
0799 
3286 
5709 
8219 
0664 
3og6 

55i4 
7918 
o3io 
2688 
5o54 
7406 
9746 
2074 
4389 
6692 

8982 
1261 
3527 
5782 
8026 
0257 
2478 
4687 
6884 
9071 



3oio3o 
3196 
53 5 1 

71'/' 

96^0 

;3n-~>; 

3-67 

8o63 

320146 

3222 19' 

65 56, 
B38ol 

33o4 1 4 
2438 
4454 
6 460 
8456i 

3404441 



1247 
3412 
5566 
7710 
9843 
1 966 

6180 

8272 

3 '> i 

2426 
4 #8 
6541 

0617 
2640 
4655 
6660 
8656 
0642 



4663 ! 

7365 ! 

••5i j 

2720 

53 7 3 

8010 

o63i 

3236 

5826 

8400 

0960 
3304 
6o33 
8548 
1048 
3534 
6006 

8464 
0908 
3338 

5755 
8i58 
o548 
2925 
5290 
7641 
9980 
2 3 06 
4620 
6921 

921 1 
1488 

3753 
6007 

8249 
0480 
2699 
4907 
7 1 04 
9289 

1 i6 { 

3628 

5781 

-•/•' i 
••56 

2177 

4289 

■ 

8481 

o562 

4694 

87*7 
0819 

2842 

4856 

6860 

8855 

0841 



4934 
7634 
•3i9 

2986 
5633 
8273 
0892 

3496 
6084 

8657 

I2l5 

3757 
6285 
8799 
1297 
3 7 s 2 
6252 

8709 

1 1 5i 

358o 

5996 
83 9 8 
0787 

3i62 
5525 
7875 

»2l3 

2538 
485o 
71 5i 

9 43o 
1715 

3979 
6232 

8473 
0702 
2920 
5 1 27 
7323 
9507 

1681 

3844 

5996 

81 3 7 

2389 

4 1 99 
65g9 
8689 
0769 

2839 
4899 

8991 

1022 

3o44 
5o57 
7060 

9054 
1039 

3 



5:o4 
7904 

•586 

3252 

5go2 
8536 
1 1 53 
3755 
6342 
8913 

U70 
40 1 1 
653 7 
9049 
1 546 
4o3o 

6499 
8934 
1395 
3822 

6237 

863] 

1025 

3399 

5761 
8110 
•446 

2770 

5o8i 
738o 

9667 
1942 
42o5 
6456 
8696 
0925 
3i4i 
5347 
7542 
972D 

1898 
40)9 
621 1 
835 1 

•4 s 1 
2600 
4710 

8898 

0977 

3 04 6 
5io5 
7 1 55 
9194 

12 23 

3246 
52 57 
7260 
9253 
1237 



5475 
Sl73 

•853 
35i8 
6166 
8798 
1414 
40 1 5 
6600 
9170 

1724 
4264 
6789 
9299 
1793 
4277 
6745 
9198 
1638 
4064 

6477 

8877 
1263 
3636 
5996 

8344 
•679 
3ooi 
53n 
7609 

9895 
2 1 69 
443 1 
6681 
8920 
1 147 
3363 
5567 
7761 
9943 

2114 

4273 
6423 
8564 
•6 9 3 
281; 
4920 
7018 
9106 
1 184 

3232 

53io 

-339 

9 $98 

1427 

5458 

7459 
945 i 
1435 



8 



5746 

8441 
1 1 2 1 

3783 
643o 
9060 

i6 7 5 

4274 
6858 
9426 

1979 
45i7 
7041 
955o 
2044 
4525 
6991 
9443 
1881 
43o6 

6718 
91 16 
i5oi 
3873 
6232 
85 7 8 
•912 
3233 
5542 
7 838 

•123 

2396 
4636 
6905 
9143 
1369 
3584 
5787 

7979 
•161 

233i 

4491 
6639 
8778 
•906 
3o23 
5i3o 
7227 

93i4 
1391 

3458 
55 1 6 
7563 
9601 
i63o 
3649 
5658 
7 65 9 
9630 
i632 



6016 
8710 
1388 
4049 
6694 
9323 
io36 
4333 
7 1 1 5 
9682 

2234 

4770 
7292 

9800 
2293 

4772 
7237 

9687 

2125 

4548 

6q58 
9355 
1739 
4109 

6467 
8812 

1 144 

3464 

5772 
8067 

•35i 
2622 
4882 
7i3o 
9 366 
1691 
38o4 
6007 
8198 
•378 

2547 
4706 

6«54 
899. 
1 1 18 
3234 
534o 
7436 
g522 
1 5 9 8 

3665 
5721 
7767 
980D 
1 83 2 
385o 
5839 
7868 

9849 

i83o 



6286 

8979 
1 654 
43i4 
6957 
9385 
2196 

4792 
7372 
9 9 38 

2488 
5o23 
7544 
•®5o 
2541 
5019 
7482 

9932 
2368 
4790 

7198 
9 5 9 4 
1076 
4346 
6702 
9046 
1377 
36 9 6 
6002 
8296 

•578 
2849 
5 1 07 
7354 
9589 
i8i3 
4025 
6226 
8416 
•5g5 

2764 
4921 
7008 
9204 
i33o 

3443 
555i 
7646 
9730 
i8o5 

38 7 i 
5926 
7972 
•••8 
2o34 
4o5i 
6059 
8o58 

••47 
2028 

8 



D. 



6556 

9247 
1921 

4579 
7221 

9846 
2456 
5o5i 
763o 
•193 

2742 
5276 

779 5 
•3oo 

2790 

5266 

7728 
•176 
2610 
5o3i 

7439 
9 833 

2214 

43S2 
6937 
9279 
1609 
3927 
6232 

8525 

•806 
3075 
5332 
7 5 7 8 
9812 
2o34 
4246 
6446 
8635 
•8i3 

29S0 
5i36 

7282 

9 • l 7 
1 '42 

3656 

7854 
99 38 

201 2 

407 

6i3i 

8176 

•.mi 

2236 
6260 

8257 

•2 ,0 

2223 



271 
269 
267 
266 
264 
262 
26l 
259 

258 
256 

254 
2531 

252 

25o 

249 

us 
246 
245 

243 

242 

241 

23 9 
238 

23 7 . 

235 
234 
233 

232 
230 
229 

228 
227 
226 
225 
223 
222 
221 
220 
219 
218 

217 
2l6 
2 I 5 
2l3 
212 
211 
210 
20O 
200 
207 

2 (16 
204 

202 

202 
201 
20O 
190 
I98 

D. 



A TABLE OF LOGAKITHMS FEOM 



1 TO 10,000 



1ST.. 

220 





1 


2 


3 


4 


5 


6 


7 


8 


9 


197 


342423 


2620 


2817 


3oi4 


3212 


3409 


36o6 


38o2 


3999 


4196 


221 


4392 


458 9 


4785 


4981 


5178 


5374 


5370 


5 7 66 


5962 


6157 


196 


222 


6333 


6549 


6744 


6 9 3 9 


7l35 


733o 


7525 


7720 


79 i5 


8110 


195 


223 


83o5 


85oo 


8694 


8889 


9 o83 


9278 


9472 


9666 


9860 


••54 


194 


224 


350248 


0442 


o636 


0829 


1023 


1216 


1410 


i6o3 


1796 


1989 


t 9 3 


225 


2i83 


2375 


2568 


2761 


2954 


3i47 


333 9 


3532 


3724 


3916 


i 9 3 


: 226 


4108 


43oi 


44g3 


4685 


48/6 


5o68 


5260 


5452 


5643 


5834 


192 


i 227 


6026 


6217 


6408 


6399 


6790 


6981 


7172 


7363 


7554 


7744 


191 


' 228 


7935 


8i25 


83i6 


85o6 


8696 


8886 


9076 


9266 


9456 


9646 


190 


229 


9&35 


•®25 


•2*5 


•404 


®5 9 3 


•783 


•972 


1161 


i35o 


i53g 


189 


23o 


361728 


1917 


2io5 


2294 


2482 


2671 


285 9 


3o48 


3236 


3424 


188 


23l 


36i2 


38oo 


3988 


4176 


4363 


455i 


4739 


4926 


5u3 


53oi 


188 


232 


5488 


56 7 5 


5862 


6049 


6236 


6423 


6610 


6796 


6 9 83 


7169 


187 


233 


7356 


7542 


7729 


7 9 i5 


8101 


8287 


8473 


863 9 


8845 


go3o 


186 


234 


9216 


9401 


9 58 7 


9772 


9958 


•143 


•328 


•5i3 


•698 


•883 


i85 


235 


371068 


1253 


1437 


1622 


1806 


1991 


2173 


236o 


2544 


2728 


184 


236 


1 

2912 


•3096 


3280 


3464 


3647 


383i 


40 1 5 


4198 


4382 


4565 


184 


237 


4748 


4932 


5u5 


5298 


5481 


5664 


5846 


6029 


6212 


63 9 4 


1 83 


238 


6577 


6759 


6942 


7124 


7306 


7488 


7670 


7852 


8o34 


8216 


182 


239 


83 9 8 


858o 


8761 


8943 


9124 


93o6 


9487 


9668 


9849 


••3o 


181 


240 


380211 


o3g2 


0573 


0754 


0934 


1 1 15 


1296 


U76 


1 656 


i83 7 


181 


241 


2017 


2197 


2377 


2657 


2737 


2917 


3o 97 


3277 


3456 


3636 


180 


242 


38i5 


3995 


4i74 


4353 


4533 


4712 


4891 


5070 


5249 


5428 


179 


243 


56o6 


5785 


5964 


6142 


6321 


6499 


6677 


6856 


7°34 


7212 


178 


144 


7390 


7D68 


7746 


79 23 


8101 


8279 


8456 


8634 


881 1 


8989 


178 


i45 


9166 


9343 


9^20 


9698 


9 8 7 5 


••31 


•228 


•4o5 


•582 


• 7 5 9 


177 


246 


390935 


1 1 1 2 


1288 


1464 


1641 


1817 


i 99 3 


2169 


2345 


2521 


176 


247 


2697 


2873 


3o48 


3224 


34oo 


3575 


3731 


3926 


4101 


4277 


176 


248 


4452 


4627 


4802 


4977 


5i52 


5326 


55oi 


5676 


585o 


6025 


i 7 5 


249 


6199 


63 7 4 


6548 


6722 


6896 


707I 


7245 


74i9 


7592 


7766 


174 


25o 


397940 


8114 


8287 


8461 


8634 


8808 


8981 


9154 


9328 


95oi 


i 7 3 


25l 


9674 


9847 


««2o 


•192 


•365 


•538 


•711 


•883 


io56 


1228 


i 7 3 


252 


40140: 


i5 7 3 


1745 


1917 


2089 


2261 


2433 


26o5 


2777 


2949 


172 


253 


3l2I 


3292 


3464 


3635 


3807 


3 97 8 


4U9 


4320 


4492 


4663 


171 


234 


4834 


5oo5 


5l76 


5346 


55i7 


5688 


5858 


6029 


6199 


6370 


171 


255 


654o 


6710 


688l 


7o5i 


7221 


73 9 i 


756i 


773i 


7901 


8070 


170 


256 


8240 


8410 


85 79 


8749 


8918 


9087 


9 25 7 


9426 


9595 


9764 


169 


257 


9933 


*I02 


©271 


®44o 


•609 


•777 


•946 


1 1 14 


1283 


U5i 


**9 


258 


411620 


I788 


1936 


2124 


229J 


2461 


2629 


2796 


2964 


3i32 


168 


. 259 


33oo 


3467 


3635 


38o3 


3970 


4i37 


43o5 


4472 


4639 


4806 


167 


260 


4i4973 


5i4o 


53o7 


5474 


564i 


58o8 


5974 


6141 


63o8 


6474 


167 


261 


6641 


6807 


6973 


7i3g 


7306 


7472 


7638 


7804 


7970 


8i35 


166 


262 


83oi 


8467 


8633 


8798 


8964 


9129 


9 2 9 5 


9460 


9625 


979i 


i65 


263 


99D6 


®i 21 


•286 


•45 1 


•616 


•781 


•945 


IIIO 


1275 


1439 


i65 


264 


421604 


1788 


1933 


2097 


2261 


2426 


23go 


2754 


2918 


3082 


164 


265 


3246 


3410 


35 7 4 


3 7 3 7 


3901 


4o65 


4228 


4392 


4355 


4718 


164 


266 


4882 


5o45 


5208 


5371 


5534 


56 9 7 


586o 


6023 


6186 


6349 


1 63 


267 
268 


65 1 1 
8i35 


6674 
8297 


6836 
845o 


6999 
8621 


7161 
8 7 83 


7324 
8944 


7486 
9106 


7648 
9268 


7811 
9429 


7973 
9391 


162 
162 


269 


97 52 


9914 


•©t5 


•236 


•3 9 8 


•559 


•720 


•881 


1042 


1203 


161! 

1 


270 


43 1 364 


i525 


i685 


1846 


2007 


2167 


2328 


2488 


2649 


2809 


161 


271 


2969 


3i3o 


3290 


345o 


36io 


3770 


3 9 3o 


4090 


4249 


4409 


160 


272 


4569 


4729 


4888 


5o48 


5207 


536 7 


5526 


56S5 


5844 


6004 


1 5 9 


2/3 


6i63 


6322 


6481 


6640 


6798 


6957 


7116 


7275 


7433 


7592 


i5o 


274 


7751 


7909 


8067 


8226 


8384 


8542 


8701 


885 9 


9017 


9173 


1 58 


275 


9 333 


9491 


9648 


9806 


9964 


•122 


• 279 


•437 


•594 


•752 


1 58 


276 


440909 


1066 


1224 


i38i 


1 538 


1695 


i852 


2009 


2166 


2323 


i5 7 


277 


2480 


2637 


2793 


2950 


3 1 06 


3263 


34i9 


3576 


3732 


3889 


1 5 7 


278 


4o45 


4201 


4337 


43i3 


4669 


4825 


49S1 


5i37 


5293 


5449 


1 56 


279 


56o4 


5760 


5915 


! 6071 


6226 


6382 


6537 


6692 
7 


6848 


7003 


i55 
D. 


N. 





1 


2 


3 


4 


5 


6 


8 


9 



A TABLE OF LOGARITHMS FROM 1 TO 10,000. 



N. 

280 
281 
282 
283 
284 
:85 
286 
287 
288 
289 

290 
291 
292 
293 

294 
29'j 
296 

2 9< 
298 

299 

3oo 
3oi 
3o2 
3o3 
3o4 
3o5 
3o6 
3o7 
3 08 
309 

3io 
3u 

3l2 

3i3 

3 14 

3i5 
3i6 
3.7 
3i8 
319 

320 
321 
322 

323 
324 
325 
326 
327 

32rf 

329 



447153 
8706 

450249 

1786 

33i8 
4845 
6366 
7882 
9392 
460898 

462398 
38 9 3 
5383 
6868 
834" 
9822 

471292 
27") 6 
4216 
5671 

477121 
8566 

480007 
1443 

2S74 
43oo 
5721 
7 i3S 
855 1 
99 58 

49i362 

2-60 
4i55 

5544 
6930 
83u 
9687 
5oio59 
2427 
3 79 i 

5o5i5o 
65o5 
7856 

9203 
5io:")/i'') 

I SSI 

3218 
4548 
5874 

7196 



33o 

33i 

332 

333 

334 

335 

336 

33 7 

338 

339 

N. 



5 1 85 1 4 

9828 

&2ii38 

2444 
3 7 //> 
5o45 
6339 
63o 



917 
53o2oo 



7 3i3 
8861 
o4o3 
1940 

3471 
4997 
65i8 

8o33 
9543 
1048 

2548 
4042 
5532 
7016 
84q5 
9969 
1438 
2go3 
4362 
58i6 

7266 
871 1 
oi5i 
1 586 
3oi6 
4442 
5863 
7280 
8692 
••99 

i5c2 
2900 

4294 

5683 
7068 
8448 
9824 
1 196 
2564 
3927 

5286 
6640 
7991 

9 337 
0679 
2017 
335 1 
4681 
6006 
7328 

8646 
9969 

1269 
2573 
3876 
5174 
6469 

77 5 9 
904^ 
o328 



3 



4 



7468 
90 1 5 
o557 
2093 
3624 
5i5o 
6670 
8184 
9694 
1198 

2697 
4191 

563o 

7164 
8643 
•116 
1 585 

3o49 
45o8 
5962 

741 1 
8855 
0294 
1729 
3 1 59 
4585 
6oo5 
742i 
8833 
•239 

1642 
3o/ t o 
4433 

5822 

7206 
8586 
9962 
i333 
2700 
4o63 

5421 
6776 
8126 

947 1 
08 1 3 
2 1 5 1 

3484 
481 5 

6139 
7460 

8777 
••90 
1400 
27o5 
4006 
53o4 
6 598 
7888 
9H4 
0456 



7623 
9170 
07 1 1 

2247 

3777 
53o2 

6821 
8336 
9845 
1 348 

2847 
4340 
5829 
7312 
8790 
•263 
1732 
3195 
4653 
6107 

7555 
8999 
0438 
1872 
33o2 

4727 
6147 

7563 

8o-4 
•38o 

1782 
3179 
4572 
5960 
7'344 
8724 
••99 

1470 
2837 
4199 

5557 
691 1 
8260 
9606 

0947 
2284 

3617 
4940 

6271 
7592 

8909 
•221 
l53o 

2835 
4 1 36 

5434 
6727 
8016 
9302 
o584 



7778 
9324 
o865 
2400 
3g3o 
5454 
6973 
8487 

999 D 
1499 

2997 

4490 

5 977 
746o 

8 9 38 

•410 

1878 

334i 

4-99 

62J2 

77OO 
9143 

o582 
2016 
3445 
4869 
6289 
7704 
91 14 

•520 

1922 

3319 

47 II 
6099 
7483 
8862 
•236 
1607 
2973 
4335 

56 9 3 
7046 
83 9 5 
974o 
1 08 1 
2418 
375o 
5079 
64o3 
7724 

9040 
•353 

1 66 1 
2966 
4266 
5563 
6856 
814J 
943o 
0712 



79 33 

9478 
1018 
2553 
4082 
56o6 
7125 
8638 
•146 
1649 

3i46 
4639 
6126 
7608 
9085 
•557 

2025 
3487 
4944 

6397 

7844 
9287 
0725 
2 1 09 
3587 
5oii 
643o 
7845 
9255 
•661 

2062 
3458 
485o 
6238 
7621 
8999 
•374 
1744 
•3109 

4471 

5828 
7181 
853o 
9874 
1 2 1 5 
2 5 5 1 
3883 
52 1 1 
6535 
] 1 i 

9171 

•484 

1792 

3096 

43 9 6 
56o3 

8274 
9 5:'> 9 
0840 



8088 
9 633 
1172 
2706 
4235 
5758 
7276 
8789 
•296 
1799 

3296 
4788 
6274 
77 56 
9233 
•704 
2171 
3633 
5090 
6542 

7989 
9 43 1 
0869 

2302 
373o 

5i53 
6572 
79S6 
9396 
•801 

2201 

3597 

4989 
6376 

7759 
9' 3 7 
•5 1 1 
1880 
3246 
4607 

5o64 

7J16 

8664 
•••9 

1 349 
2684 
4016 
5344 
6668 

7987 
9 3o3 
•6i5 
1922 
3226 
4526 
5822 
711 i 
8402 
9687 
0968 



8 



8242 

9787 
i326 
2859 
438 7 
5910 
7428 
8940 

•447 
1948 

3445 
4936 
6423 
7904 
Q 38o 
•85i 
2 3i8 

3770 
5235 
6687 

8i33 
9 5 7 5 
1012 
2445 
3872 
5295 
67U 
8127 
9537 
•941 

2341 

3737 
5i28 
65i5 

7897 
9 2 7 5 

•648 

2017 

3382 

4743 

6099 
745i 

8799 
•143 
1482 
2818 

4149 

5176 

6800 
8119 

9434 
•745 
2o53 
3356 
4656 
5951 
7243 
853i 
9815 
1096 



83 97 I 

994i 

1479 
3oi2 

4540 
6062 

7579 
9091 
•597 

2098 

35 9 4 
5o85 
6571 
8o52 
9527 
•998 
2464 
3g25 
538i 
6832 

8278 

97'9 
n56 

2583 

40 1 5 

5437 
6855 
8269 

9 6 77 
1081 

2481 
38 7 6 
5267 
6653 
8o35 
9412 
•785 
2i54 
35i8 
4878 

6234 
7 586 
8 9 34 
•277 
1616 
2g5i 
4282 
5609 
6932 
825i 

9 566 
•876 
2 1 S3 
3486 

9943 

1223 



8552 
••95 
1633 
3i65 
4692 
6214 
7 7 3i 
9242 
•748 
2248 

3 7 44 
5234 
6719 
8200 
9675 
ii45 
2610 
4071 
5526 
6976 

8422 
9 863 
1299 
2731 
4i57 
55 79 
6997 
8410 
9818 
1222 

2621 
40 1 5 
54o6 
6791 
8173 
955o 
•922 
2291 
36d5 
5oi4 

6370 
7721 
9068 
•4 1 1 
1750 
3o84 
44i4 
5741 
7064 
8382 

9697 
1007 

23 1 4 
36i6 

49 1 5 
6210 
t5oi 

••72 
i35i 



D. \ 

i 

1 55 

1 54 
i54 
l53 
i53 
i5a 

l52 

i5i 
i5i 
i5o 

i5o 
149 

1491 

148 I 
148 

147! 
146 i 
146 i 
146 I 
145 

145 1 

144 

U4 

143 

143 

142 

142 

141 

141 

140 

140 

1 39 

139 

1 3c 

i3t 

i38 

i3 7 

i3 7 

i36 

i36 

1 36 
i35 
1 35 
1 34 
1 34 
1 33 
1 33 
1 i I 

I 32 

i3a 

i3i 

i3i 
i3i 
i3o 
i3o 

129 
129 

12c 

1 at 

128 



6 



A TABLE OF LOGARITHMS FROM 1 TO 10,000. 



N. 
34o 





1 


2 


3 


4 


5 


6 


7 


8 


9 


D. 


531479 


1607 


1734 


| 1862 


1990 


2 1 1 7 


2245 


2372 


25oo 


2627 


128 


34i 


2754 


2882 


3009 


1 3i36 


3264 


3391 


35i8 


3645 


3772 


38 99 


127 


342 


4026 


4i53 


4280 


4407 


4534 


4661 


4787 


4914 


5o4i 


5ib~i 


127 


343 


5294 


5421 


1 5547 


56 7 4 


58oo 


5927 


6o53 


6180 


63o6 


6432 


126 


344 


6 558 


66b5 


681 1 


6937 


7063 


7189 


73x5 


7441 


7567 


7693 


126 


34 J 


7819 


' 7945 


8071 


8197 


8322 


8448 


85 7 4 


8699 


6825 


8931 


126 


346 


9076 


9202 


9 32 7 


9452 


9 5 7 8 


97 o3 


9829 


9954 


*®79 


•204 


125 


347 


540^29 


o455 


o58o 


0705 


o83o 


0955 


1080 


1205 


i33o 


1454 


125 


34« 


i5 79 


1704 


1829 


1953 


2078 


2203 


2327 


2452 


2576 


2701 


125 


349 


28i5 


2 9 5o 


3e 74 


3199 


3323 


3447 


3571 


3696 


3320 


3944 


124 


35o 


544068 


4192 


43i6 


4440 


4564 


46S8 


4812 


4936 


5o6o 


5i83 


124 


35i 


5307 


543 1 


5555 


56 7 8 


58o2 


5925 


6049 


6172 


6296 


6419 


124 


352 


6543 


6566 


6789 


6913 


7o36 


7i5 9 


7282 


74o5 


7629 


7652 


123 


353 


7775 


7898 


bo2l 


8144 


8267 


838 9 


85i2 


8635 


8 7 58 


888 1 


123 


, 354 


9003 


9126 


9249 


9371 


9494 


9616 


9 7 3 9 


9861 


9984 


•106 


123 


355 


550228 


o35i 


0473 


0595 


0717 


0840 


0962 


1084 


1206 


1328 


122 


356 


i45o 


1572 


1694 


1816 


ig38 


2060 


2181 


23o3 


2420 


2547 


122 


357 


2668 


2790 


2911 


3o33 


3i55 


3276 


33 9 8 


35i9 


364o 


3 7 62 


121 


353 


3883 


4004 


4126 


4247 


4368 


4489 


4610 


473i 


4852 


4973 


121 


359 


5og4 


52ID 


5336 


5457 


5578 


5699 


5820 


5g40 


6061 


6182 


121 


36o 


5563o3 


6423 


6544 


6664 


6785 


6905 


7026 


7146 


7267 


7 38 7 


120 


36 1 


7507 


7627 


7748 


7868 


7988 


8108 


8228 


8349 


8469 


8589 


120 


362 


8709 


8829 


8948 


9068 


9188 


93o8 


9428 


9548 


9667 


9787 


120 


363 


9907 


®» 2 6 


•146 


•265 


•385 


•5o4 


•624 


•743 


•863 


•982 


119 


364 


56noi 


1221 


1 34o 


1459 


i5 7 8 


1698 


1817 


1936 


2o55 


2174 


II9 


365 


2293 


2412 


253i 


265o 


2769 


2S87 


3oo6 


3i 2 5 


3244 


3362 


119 


366 


348i 


36oo 


3 7 i8 


3837 


3955 


4074 


4192 


43n 


4429 


4543 


II9 


*67 


4666 


4784 


4903 


5o2i 


5i39 


5257 


5376 


5494 


56i2 


5730 


Il8 


J68 


5848 


5966 


6084 


6202 


6320 


6437 


6555 


66 7 3 


679 1 


6909 


Il8 


J69 


7026 


7*44 


7262 


7 3 79 


7497 


7614 


7 7 32 


7849 


7967 


8084 


Il8 


{70 


568202 


83i 9 


8436 


8554 


8671 


8788 


8 9 o5 


9023 


9140 


9257 


117 


»7' 


9374 


9491 


9608 


9725 


9842 


99 5 9 


•976 


•io3 


•309 


•426 


117 


Sl \ 


D70543 


0660 


0776 


0893 


IOIO 


1126 


1243 


13D9 


1476 


1592 


II 7 


J 7 3 


1709 


1825 


1942 


2o58 


2174 


2291 


2407 


2523 


2639 


2735 


Il6 


3 7 4 


2872 


2988 


3 1 04 


3220 


3336 


34D2 


3563 


3684 


3800 


3915 


ll6 


373 


4o3i 


4147 


4263 


43 79 


4494 


4610 


4726 


4841 


49 5 7 


0072 


Il6 


3 7 6 


5i88 


53o3 


5419 


5534 


565o 


5765 


588o 


5996 


61 1 1 


6226 


n5 


3 7 Z 


634i 


6457 


6572 


6687 


6802 


6917 


7032 


7U7 


7262 


7377 


u5 


378 


7492 


7607 


7722 


7836 


79 5i 


8066 


8181 


8295 


84io 


S525 


ii5 


379 


863 9 


8754 


8868 


8 9 83 


9097 


9212 


9326 


9441 


q555 


9669 


H4 


38o 


579784 


9898 


••12 


•126 


®24i 


•355 


•469 


•583 


•697 


•811 


114 


38i 


580925 


io39 


1 1 53 


1267 


i38x 


1495 


1608 


1722 


i836 


i 9 5o 


114 


382 


2o63 


2177 


2291 


2404 


25i8 


263 1 


2745 


2858 


2972 


3o35 


114 


383 


3199 


33 1 2 


3426 


3539 


3652 


3765 


3879 


3 99 2 


4io5 


4218 


n3 


384 


433 1 


4444 


4557 


4670 


4783 


4896 


5009 


5l22 


5235 


5348 


n3 


385 


D461 


55 7 4 


5686 


5 799 


5912 


6024 


6137 


6250 


6362 


6475 


n3 


386 


65o7 


6700 


6812 


6925 


7037 ' 


7*49 


7262 


7374 


7486 


7399 


112 


38 7 


7711 


7823 


7 9 35 


8047 


8160 


8272 


8334 


8496 


8608 


0720 


112 


388 


8832 


8944 


9o56 


9167 


9279 


9 3 9 i 


95o3 


96l5 


9726 


9838 


112 


389 


99J0 


••61 


♦173 


•284 


•3 9 


•507 


•619 


•730 


•842 


•953 


112 


390 


391065 


1 1 76 


1287 


1 3 99 


i5io 


it vi 


1732 


1843 


1955 


2066 


in 


3qi 


2177 


2288 


23 99 


25l0 


2621 


2732 


2843 


2954 


3o64 


3i75 


in 


392 


3286 


33 97 


35o3 


36i8 


3729 


3840 


3950 


4o6i 


4171 


4282 


in 


3 9 3 


43 9 3 


45o3 


4614 


4724 


4834 


4945 


5o55 


5i65 


5276 


5386 


no 


3 9 4 


5496 


56o6 


5717 


5827 


5 9 37 


6047 


6157 


6267 


63 7 7 


6487 


no 


395 


6597 


6707 


6817 


6927 


^037 


7146 


7256 


7366 


7476 


7586 


no 


3 9 6 


7695 


78o5 


79'4 


8024 


8i34 


8243 


8353 


8462 


8572 


8681 


no 


397 


8791 


8900 


9009 


9119 


9228 


9337 


9446 


9556 


9665 


9774 


109 


398 


9 883 


9992 


•101 


•210 


•319 


•428 


•537 


•646 


•755 


•864 


109 


399 


600973 


1082 


1191 


1299 


1408 


i5i7 


i625 


1734 


1843 


1951 


109 


N. 





1 


a 


3 


4 


5 


6 


7 


8 


9 



A TABLE OF LOGARITHMS FROM 1 TO 10,000. 



IT. 





1 


2 


3 
* 


4 


5 


6 


7 


8 


9 


D. 


400 


602060 


2169 


2277 


2386 


2494 


2603 


2711 


2819 


2928 


3o36 


108 


401 


3144 


3 2 53 


336z 


3469 


35 77 


3686 


37Q4 


3902 


4010 


4n8 i 


108 


402 


4220 


4334 


4442 


455o 


4658 


4766 


4874 


49^2 


D089 


5197 


108 


4c3 


53o5 


5;i 3 


552i 


D628 


5i36 


5^44 


D951 


6059 


6166 


6 Vi 


108 


404 


63 m 


6489 


6096 


6704 


681 1 


6919 


-026 


7i33 


7241 


7348 


107 


4o5 


7455 


7062 


7609 


7777 


7»84 


7991 


8098 


82o5 


83i2 


8419 


107 


406 


8026 


bo33 


8740 


8S47 


8954 


9061 


9167 


9 2 74 


9381 


94H8 


107 


407 


9)94 


9701 


9808 


9914 


••21 


•128 


•234 


Gj4i 


•447 


•554 


107 


4o£ 


6 1 0660 


0767 


0873 


0979 


1086 


1192 


129b 


i4o5 


i5n 


1617 


106 


409 


i 7 23 


1S29 


1936 


2042 


2148 


2254 


236o 


2466 


2572 


2078 


106 


410 


(12784' 


2S90 


2996 


3l02 


3207 


33i3 


3419 


3525 


363o 


3736 


106 


411 


3842 


3947 


4o53 


4i59 


4264 


4370 


4475 


458i 


4686 


4792 


106 


412 


4^97 


5oo3 


5io8 


02l3 


5319 


5424 


5529 


5634 


5740 


5845 


io5 


4i3 


5q5j 


6o55 


6160 


6265 


6070 


6476 


658i 


6686 


6790 


68 9 5 


jo5 


4i4 


7000 


7io5 


7210 
8257 


73i5 


7420 


7325 


7629 


7734 


7 83 9 


7943 


io5 


4i5 


8048 


8i53 


8362 


8466 


8571 


8676 


8780 


8884 


8989 


io5 


416 


9093 


9198 


9302 


9406 


9D1 1 


9615 


9719 


9824 


9928 


••32 


104 


4i7 


620136 


0240 


o344 


0448 


o552 


o656 


0760 


0S64 


0968 


1072 


104 


418 


1 176 


1 2 So 


1384 


1488 


1592 


1695 


H99 


1903 


2007 


2110 


104 


419 


2214 


23 1 a 


2421 


2525 


2628 


2732 


2835 


2939 


3o42 


3i46 


104 


420 


623249 


3353 


3456 


3559 


3663 


3766 


386 9 


3973 


4076 


4179 


io3 


421 


4282 


4385 


4488 


4591 


4695 


4793 


4901 


5oo4 


5107 


5210 


io3 


422 


53 1 2 


54i5 


55 1 8 


562i 


5724 


5S27 


5929 


6o32 


6i35 


6238 


io3 


423 


634o 


6443 


6546 


6648 


6751 


6853 


6956 


7o58 


7161 


7263 


io3 


424 


7366 


7468 


7571 


7673 


777 5 


7878 


7980 


8082 


8*85 


8287 


102 j 


423 


838g 


8491 


8593 


8695 


8797 


8900 


9002 


9104 


9206 


93o8 


102 


426 


94 io 


95 1 2 


9613 


97i5 


9817 


9919 


••21 


•123 


•224 


•326 


102 


427 


63o428 


o53o 


o63i 


0733 


08 3 5 


0936 


io38 


1 139 


1241 


i342 


102 


428 


1444 


1 545 


1647 


1748 


1849 


1951 


2052 


2i53 


2255 


2356 


101 


429 


2457 


2559 


2660 


2761 


2862 


2963 


3064 


3i65 


3266 


3367 


101 


43o 


633468 


356g 


3670 


3771 


38 7 2 


3 97 3 


4074 


4n5 


4276 


4376 


100 


43 1 


4477 


4578 


4679 


4779 


4880 


4981 


5o8i 


5182 


5283 


5383 


100 


432 


5484 


5584 


568o 


57»5 


5886 


5986 


6087 


6187 


6287 


6388 


100 


433 


6488 




6688 


6789 


6889 


6989 


7089 


7189 


7290 


73oo 


100 


434 


7490 


7590 


7690 


7790 


789° 


7990 


8090 


8190 


8290 


838g 


99 


435 


8489 


8589 


86*9 


87*9 


8888 


8 9 58 


9088 


9188 


9287 


9387 


99 


436 


9/166 


9 586 


9686 


9783 


9*85 


9984 


••84 


•i83 


•283 


•382 


99 


437 


640481 


o58i 


0680 


°779 


0879 


0978 


1077 


1177 


1276 


1373 


99 


438 


i474 


1 - ; 


1672 


1771 


1871 


1970 


2069 


2168 


2267 


2366 


99 


439 


2 465 


2 563 


2662 


2761 


2S60 


2939 


3o58 


3 1 56 


3255 


3354 


99 


440 


643453 


355 1 


365o 


3749 


3847 


3946 


4044 


4i43 


4242 


4340 


9 8 


441 


44 5 9 


4537 


4636 


4734 


4832 


493i 


5029 


5127 


5226 


5324 


98 


442 


5422 


5521 


56 1 9 


5717 


58i5 


59 1 3 


601 1 


61 10 


6208 


63o6 


98 


443 


64 >>/, 




6600 


6698 


6796 


6H94 


6992 


7089 


7187 


7285 


98 


444 


7383 




7)79 


7676 


7774 


7872 
8848 


7969 


8067 


8i65 


8262 


98 


44 5 


836o 




85 >5 


8653 


8750 


8945 


9043 


9140 


9237 


97 


446 


9JJ5 


943a 


9 >3o 


9627 


9724 


9821 


99'9 


••16 


•n3 


•210 


97 


44/ 


65o3o8 




OJ02 


0599 


0696 


0193 


0800 


0987 


1084 


1181 


97 


448 


1278 


ii-) 


1472 


1569 


1660 


1762 


i85o 


1956 


2o53 


2i5o 


97 


449 


2246 




2440 


2536 


2633 


2730 


2S26 


2923 


3oig 


3u6 


9" 


i5i 


6532 1 3 


3309 


34o5 


35o2 


35 9 8 


36o5 


3791 
4754 


3888 


3 9 84 


4080 


96 


*5i 


4'77 


42-; 


4 I69 


4465 


4562 


4658 


485o 


4946 


5o4a 


96 


452 


5i38 


52 J 5 


533 1 


5427 


5523 


5619 


5 7 i5 


58io 


5906 
6864 


6002 


96 


453 


609S 6194 
7o56 7162 


()2(jO 


6386 


648a 


6377 


66 7 3 


6769 




90 


454 


7247 


7343 


7438 


7534 

8488 


7629 
8584 


7723 


7S20 


7916 


96 


455 


8011 8107 


820 2 


»2<,S 


83g3 


8679 


8774 




9 * 


456 


8965 9060 


g i 5 > 


9230 


9346 


944i 


(/) ](> 


96 3 1 


9726 


9s 21 


9 5 


457 


9916 ## ii 


•106 


•201 


•296 


•39' 
1JJ9 


•486 


•58i 


•676 


•77' 


n 


458 


660^6 5 0960 


io55 


ii5o 


1245 


1434 


:52o 


1623 


1718 


95 


459 


i8i3 1907 


2002 


2096 


2191 


2286 


238o 


247^ 


2569 
8 




95 

D. 


N. 





1 


2 


3 


4 


5 

1 


6 

— 


1 


9 



L TABLE OF LOGARITHMS FROM 1 TO 10, 000. 



N. 





1 


1 
2 | 


3 


4 


5 


6 


7 1 


8 


9 ' 


D. 


460 


3627SS 


2852 


, 1 
2947 


3o4i 


3i35 


323o 


3324 


3418 


35i2 


3607 

— A 


94 


461 


3701 


3795 


3889 


3 9 83 


4078 


4172 


4266 


436o 


4454 


4548 


94 


462 


4642 


4736 


483o 


4924 


5oi8 


5lI2 


5206 


5299 


53 9 3 


5487 


94 


463 


558i 


5675 


5769 


5862 


5 9 56 


6o5o 


6i43 


6237 


633 1 


6424 


94 


464 


65i8 


$612 


6 7 o5 


6799 


6892 


6986 


7079 


7173 


7266 


7360 


94 


465 


7453 


7546 


7640 


7733 


7826 


7920 


8oi3 


8106 


8199 


8293 


90 


466 


8386 


8479 


8372 


8665 


8 7 5 9 


8852 


8945 


9 o38 


9i3i 


9224 


9 ? 


467 
460 


9317 


9410 


93o3 


9596 


9689 


9782 


9 8 7 5 


9967 


••60 


•i53 


9 3 


670246 


0339 


043 1 


o524 


0617 


0710 


0802 


0893 


0988 


1080 


93 


469 


1 173 


1265 


i358 


1431 


1 543 


i636 


1728 


1821 


1913 


2003 


9 3 


470 


672098 

3021 


2190 


2283 


2373 


2467 


256o 


2632 


2744 


2836 


2929 


92 


471 


3n3 


32o5 


32 97 


3390 


3482 


35 7 4 


3666 


3758 


385.0 


92 


472 


3942 

4861 


4o34 


4126 


4218 


43io 


4402 


4494 


4586 


46_77 


4769 


92 


473 


4953 


5o45 


5i3 7 


5228 


5320 


5412 


55o3 


5595 


568 7 


92 


474 


5778 


5870 


5962 


6o53 


6i45 


6236 


6328 


6419 


65n 


6602 


9 2 


475 


6694 


6 7 85 


6^76 


6968 


7o5 9 


7 i5i 


7242 


7333 


7424 


75i6 


9 1 


476 


7607 


7698 


7789 


7881 


7972 


8o63 


8i54 


8245 


8336 


8427 


9* 


477 


85i8 


8609 


8700 


8791 


8882 


8 97 3 


9064 


9i55 


9246 


9337 


9i 


47^ 


9428 


9 5i 9 


9610 


9700 


979 1 


9882 


997 3 


••63 


•i54 


•245 


9 1 


! 479 


68o336 


0426 


o5i7 


0607 


0698 


0789 


0879 


0970 


1060 


ii5i 


9 1 


4S0 


681241 


i332 


1422 


i5i3 


i6o3 


1693 


1784 


1874 


1964 


2o55 


90 


481 


2i45 


2235 


2326 


2416 


25o6 


2596 


2686 


2777 


2867 


2 9 5 7 


90 


482 


3o47 


3i37 


3227 


33i7 


3407 


3497 


3587 


36 77 


3767 


3837 


90 


483 


3947 


4o3] 


4127 


4217 


43o7 


4396 


4486 


45 7 6 


4666 


4756 


90 


484 


4845 


4933 


5023 


5 li 4 


5204 


52q4 


5383 


5473 


5563 


5652 




485 


5742 


533i 


5o2i 


6010 


6100 


6189 


6279 


6368 


6458 


6547 


89 


486 


6636 


6726 


68i5 


6904 


6994 


7088 


1 r T) 
I 1 ," 


T)At 

; 




/44Q 


^ 


487 


7529 


7618 


77°7 


7796 


7886 


7975 


8064 


8i53 


8242 


833 1 


89 


488 


8420 


8509 


85 9 8 


8687 


8776 


8865 


8 9 53 




9i3i 


9220 


t 9 


489 


9 3o 9 


9398 


9486 


9 5 7 5 


9664 


97 53 


9841 


993o 


••19 


•107 


89 


490 


690196 
1081 


0285 


0373 


0462 


o55o 


o63g 


O728 


08l6 


0905 


0993 


89 


491 


1170 


1258 


1 347 


1435 


1524 


l6l2 


1700 


1789 


1877 


88 


492 
4g3 


1965 

2847 


2o53 


2142 


2230 


23i8 


2406 


2494 


2583 


2671 


2739 


88 


2935 


3o23 


3 1 1 1 


3199 


3287 


3375 


3463 


355i 


363 9 


88 


^7 

494 


3 7 2 7 


38i5 


3903 


3991 


4078 


4166 


4254 


4342 


443 


45i7 


88 


495 


46o5 


4693 


4781 


4868 


4g56 


5o44 


5i3i 


5219 


5307 


53 9 4 


88 


496 


5482 


5569 


5637 


5744 


5832 


5919 


6007 


6094 


6182 


6269 


8t 


1 497 


6356 


6444 


653 1 


6618 


6706 


6703 


6880 


6968 


7o55 


7142 


87 


' 498 


7229 


7317 
8188 


74o4 


7491 


7 5 7 8 


76c 5 


77 52 


7839 


7926 


8014 


87 


1 
499 


81 OS 


8275 


8302 


8449 


8535 


8622 


8709 


8796 


8883 


87 


5oo 


698970 


9 o5 7 


9144 


923i 


9317 


94o4 


9491 


9 5 7 8 


9664 


975i 


87 


5oi 


9 838 


9924 


••11 


••98 


•184 


•271 


•35S 


•444 


•53 1 


•617 


87 


502 


700704 


0790 


0877 


0963 


io5o 


n36 


1222 


1 309 


i3 9 5 


1482 


86 


5o3 


1 568 


1634 


I74i 


1827 


igi3 


1999 
2861 


2086 


2172 


2238 


2344 


86 


i 5o4 


243 1 


2517 


26o3 


2689 


2 77 5 


2947 


3o33 


3i 19 


32o5 


86 


5o5 


32oi 


33 77 


3463 


3549 


3635 


3721 


3807 


38c;3 


3979 


406 5 


86 


5o6 


41 5i 


4236 


432 2 


4408 


4494 


4579 


4665 


4731 


483 7 


4922 


86 


607 


5oo8 


5094 


5179 


5265 


535o 


5436 


5522 


5607 


56g3 


5 77 8 


86 


5o8 


5864 


5 9 4 9 


6o35 


6120 


6206 


6291 


6376 


6462 


6547 


6632 


85 


009 


671b 


68o3 


6888 


6974 


7059 


7144 


7229 


73i5 


74oo 


7485 


85 


5io 


707570 


7655 


774o 


7826 


791 1 


7996 


8081 


8166 


825i 


8336 


85 


5u 


( / ' 

8421 


85o6 


85g i 


8676 


8761 


8846 


8 9 3 1 


90i5 


9100 


9185 


85 


5l2 


927c 


9355 


9440 


9524 


9609 


9694 


9779 


9 863 


9948 


••33 


85 


5i3 


710117 


0202 


0287 


0371 


o456 


o54o 


0625 


0710 


0794 


0879 


85 


5i4 


o 9 63 


1048 


ll32 


1217 


i3oi 


1 385 


1470 


1 554 


1639 


1723 


84 


5l5 


1807 


1892 


1976 


2060 


2144 


2229 


23i3 


2397 


2481 


2566 


84 


D16 


265c 


2734 


2818 


2902 


2986 


3070 


3i54 


3238 


3323 


3407 


84 


517 
5i8 


34gi 


3575 


3559 


0742 


3826 


3910 


3994 


4078 


4162 


4246 


84 


433c 


1 44i4 


4497 


45Si 


4665 


4749 


4833 


4916 


5ooo 


5o84 


84 


^19 


5i6" 


525i 


5335 


5418 


55o2 

l 


5586 


566 9 


5753 

. 


5836 


D920 


84 

1 


IT. 





1 


2 


3 


4 


5 


6 


7 


8 






A TABLE OF LOGARITHMS FROM 1 TO 10,000. 



9 




10 



A TABLE OF LOGARITHMS FROM 1 TO 10,000. 



N. 
58o 





1 
35o3 


2 


3 


4 


5 


6 


7 


8 


9 


I>. 


763428 


35 7 8 


3653 


3727 


38o2 


38 7 7 


3952 


4027 


4101 


7 5 


58i 


4176 


425i 


4326 


44oo 


4475 


455o 


4624 


4699 


4774 


4848 


73 


582 


4923 


499 s 


5072 


5i47 


5221 


5296 


5370 


5445 


552o 


55 9 4 


75 


583 


5669 


5743 


58i8 


58 9 2 


5 9 66 


6041 


6u5 


6190 


6264 


6338 


74 


584 


64i3 


6487 


6562 


6636 


6710 


678D 


685 9 


6 9 33 


7007 


7082 


74 


585 


7106 


7230 


73o4 


7379 


7453 


7527 
8268 


7601 


7 6 7 5 


7749 


7823 


74 


586 


7898 


7972 


8046 


8120 


8194 


8342 


8416 


8490 


8564 


74 


58 7 


8638 


8712 


8786 


8860 


8934 


9008 


9082 


9i56 


923o 


93o3 


74 


588 


9 3 77 


945i 


9525 


9 5 99 


9673 


9746 


9820 


9894 


9968 


c©42 


74 


58 9 


770110 


0189 


0263 


o336 


0410 


0484 


0557 


o63i 


0705 


0778 


74 


590 


770862 


0926 


0999 


1073 


1 146 


1220 


1293 


1367 


1440 


i5i4 


74 


591 


i58 7 


1661 


1734 


1 80S 


1881 


1955 


2028 


2102 


2175 


2248 


73 


592 


2322 


2395 


2468 


2542 


26i5 


2688 


2762 


2835 


2908 


29S1 


73 


5 9 3 


3o55 


3128 


3201 


3274 


3348 


3421 


3494 


3567 


364o 


37i3 


73 


594 


3 7 86 


386o 


3 9 33 


4006 


4079 


4i52 


4225 


4298 


4371 


4444 


7 5 


593 


45i7 


4590 


4663 


4736 


4809 


4882 


4955 


5o28 


5 1 00 


5n3 


73 


5 9 6 


5246 


53i9 


5392 


5465 


5538 


56 10 


5683 


5 7 56 


5829 


5902 


73 


5 97 


5974 


6047 


6120 


6193 


6265 


6338 


641 1 


6483 


6556 


6629 


73 


5 9 8 


6701 


6774 


6846 


6919 


6992 


7064 


7i3 7 


7209 


7282 


7354 


73 


5 99 


7427 


7499 


7 5 7 2 


7644 


7717 


7789 


7862 


7 9 34 


8006 


8079 


72 


600 


778i5i 


8224 


8296 


8368 


8441 


85i3 


8585 


8658 


8730 


8802 


72 


601 


8S74 


8947 


9019 


9091 


9i63 


9236 


93o8 


938o 


9452 


9524 


72 


602 


9596 


9669 


974i 


9813 


9 885 


9957 


••29 


•101 


•173 


•245 


72 


6o3 


780317 


o38 9 


0461 


o533 


o6o5 


0677 


0749 


0821 


0893 


0965 


72 


604 


1037 


1 1 09 


1 1.81 


1253 


1324 


1396 


1468 


1540 


1612 


1684 


72 


6o5 


1755 


1827 


1899 


1971 


2042 


2114 


2186 


2258 


2329 


2401 


72 


606 


24"?3 


2544 


2616 


2688 


2759 


283i 


2902 


2974 


3o46 


3 1 17 


72 


607 


3i8 9 


3260 


3332 


34o3 


3475 


3546 


36i8 


36S9 


3 7 6i 


3832 


71 


608 


3904 


3975 


4046 


4118 


4189 


4261 


4332 


44o3 


4475 


4546 


71 


609 


4617 


4689 


4760 


483 1 


4902 


4974 


5o45 


5n6 


5187 


5259 


71 


610 


78533o 


54oi 


5472 


5543 


56i5 


5686 


5 7 5 7 


5828 


5899 


5 97 o 


71 


611 


6041 


6112 


6i83 


6254 


6325 


63 9 6 


6467 


6538 


6609 


6680 


V 


612 


6751 


6822 


68 9 3 


6964 


7o35 


7106 


7177 


7248 


7319 


73 9 o 


71 


6i3 


7460 


753i 


7602 


7673 


7744 


7813 


7885 


7956 


8027 


8098 


71 


614 


8168 


8239 


83io 


838i 


8451 


8522 


85 9 3 


8663 


8734 


8804 


7> 


6i5 


8870 


8946 


9016 


9087 


9157 


9228 


9299 


9369 


9440 


95io 


71 


616 


9 5Si 


965i 


9722 


9792 


9 863 


9933 


•••4 


••74 


•144 


•2l5 


70 


617 


790285 


o356 


0426 


0496 


0567 


0637 


0707 


0778 


0848 


0918 


70 


618 


0988 


1059 


1 1 29 


1 199 


1269 


1 340 


1410 


1480 


i55o 


1620 


70 


619 


1691 


1761 


i83i 


1901 


1971 


2041 


21 11 


2181 


2252 


2322 


70 


620 


792392 


2462 


2532 


2602 


2672 


2742 


2812 


2882 


2952 


3022 


70 


621 


3o 9 2 


3i62 


323i 


33oi 


3371 


3441 


35u 


358i 


365 1 


372I 


70 


622 


3790 


386o 


3930 


4000 


4070 


4139 


4209 


4279 


4349 


44l8 


70 


'623 


4188 


4558 


4627 


4697 


4767 


4836 


4906 


4976 


5o45 


5n5 


70 


624 


5i85 


5254 


5324 


53 9 3 


5463 


5532 


56o2 


5672 


5741 


58n 


70 


623 


588o 


5 9 49 


6019 


6088 


6i58 


6227 


6297 


6366 


6436 


65o5 


69 


626 


65 7 4 


6644 


6 7 i3 


6782 


6852 


6921 


6990 


7060 


7129 


7198 


69 


627 


7268 


7 33 7 


7406 


7475 


7545 


7614 


7683 


77 52 


7821 


7890 


6 9 


628 


7960 


8029 


8098 


8167 


8236 


83o5 


83 7 4 


8443 


85i3 


8582 


69 


629 


865 1 


8720 


8789 


8858 


8927 


8996 


9065 


9i34 


9203 


9272 


69 


63o 


799341 


9409 


9478 


9547 


9616 


9 685 


9754 


9823 


9892 


9961 


69 


63 1 


S00029 


0098 


0167 


0236 


o3o5 


0373 


0442 


o5n 


o58o 


0648 


69 


632 


0717 


0786 


o854 


0923 


0992 


1061 


" 2 9 


1 198 


1266 


1335 


69 


633 


1404 


1472 


i54i 


1609 


1678 


1747 


i8id 


1884 


1952 


2021 


69 


634 


2089 


2i58 


2226 


2290 


2363 


2432 


25oo 


2568 


2637 


27o5 


n 


635 


2774 


2842 


2910 


2979 


3o47 


3i 16 


3i84 


3252 


3321 


3389 


636 


3457 


3525 


3594 


3662 


3730 


3798 


3867 
4548 


3935 


4oo3 


4071 


68 


63 7 


4139 


4208 


4276 


4344 


4412 


4480 


4616 


4685 


4753 


68 


638 


4821 


4889 


49 5 7 


5oa5 


5093 


5i6i 


5229 


5297 


5365 


5433 


68 


63 9 


55oi 


5569 


5637 


5703 


5 77 3 


584i 


5 9 o8 


5 97 6 


6o44 


6112 1 


68 


N. 





1 


1 " 


3 


4 


5 


6 


7 


8 


9 



A TABLE OF LOGARITHMS FROM 1 TO 10,000. 1! 




12 



A TABLE OF LOGARITHMS FROM 1 TO 10,000 



N. 





I 


2 


3 


4 


5 


! 6 


7 


8 


9 


■ 
D. 

(2 


700 


843098 


5 1 60 


5222 


5284 


5346 


5408 


5470 


5532 


55 9 4 


5656 


701 


5 7 i8 


,5780 


3842 


5go4 


5 9 66 


6028 


6090 


6i5i 


62i3 


6275 


62 


702 


633 7 


6399 


6461 


6523 


6385 


6646 


6708 


6770 


6832 


6894 


62 


703 


6955 


7017 


7079 


7141 


7202 


7264 


7326 


7338 


7449 


75n 


62 


704 


7573 


7634 


7696 


77 58 


7819 


7881 


7943 


8004 


8066 


81 28 


62 


700 


8189 


8201 


83i2 


83 7 4 


8435 


8497 


8539 


8620 


8682 


8i/.3 


62 


706 


88o5 


8866 


8928 


8989 


9o5i 


9112 


9174 


9235 


9297 


9 358 


61 


707 


9(19 


948i 


9342 


9604 


9665 


9726 


9788 


9849 


991 1 


9972 


61 


708 


8joo33 


0095 


oi56 


0217 


0279 


o34o 


0401 


0462 


0024 


o585 


61 


709 


0646 


0707 


0769 


o83o 


0891 


0952 


1014 


1075 


ix36 


1197 


61 


710 


85i258 


1 3 20 


i38i 


1442 


i5o3 


1 564 


1625 


1686 


H47 


1809 


61 


711 


1870 


1931 


1992 


2o53 


2114 


2175 


2236 


2297 


2358 


2419 


61 


712 


2480 


2341 


2602 


2663 


2724 


2 7 85 


2846 


2907 


2968 


3029 


61 


7 i3 


3090 


3i5o 


321 1 


3272 


3333 


33 9 4 


3455 


35i6 


3377 


3637 


61 


7i4 


36g8 


3759 


3820 


388i 


3941 


4002 


4o63 


4124 


4i85 


4245 


61 < 


7 i5 


43 06 


4367 


4428 


4488 


4549 


4610 


4670 


473i 


4792 


4852 


61 


716 


4913 


4974 


5o34 


5095 


5 1 56 


52i6 


5277 


5337 


5398 


545 9 


61 


717 


5519 


558o 


5640 


5701 


5761 


5822 


5882 


5 9 43 


6oo3 


6064 


01 


718 


6124 


6i85 


6245 


63o6 


6366 


6427 


6487 


6348 


6608 


6668 


60 


719 


6729 


6789 


685o 


6910 


6970 


7o3i 


7091 


7102 


7212 


7272 


60 


720 


857332 


7 3 9 3 


7453 


75i3 


7574 


7634 


7694 


7755 


78i5 


7875 


60 


721 


79 35 


799 D 


8o56 


8116 


8176 


8236 


8297 


8357 


8417 


8477 


60 


722 


853 7 


8397 


865 7 


8718 


8778 


8838 


8898 


8 9 58 


9018 


9078 


6(7 1 


723 


9 i38 


9198 


9258 


9 3i8 


9 3 79 


943g 


9499 


9539 


9619 


9679 


60 


724 


9739 


9799 


9859 


9918 


9978 


••38 


••98 


•i58 


•218 


®278 


60 


725 


86o338 


o3g8 


o458 


0618 


0378 


0637 


0697 


0757 


0817 


0S77 


60 


726 


oq37 


0996 


io56 


11 16 


1 1 76 


1236 


1295 


i355 


Ui5 


1475 


60 


727 


1 534 


i5 9 4 


i654 


1714 


1773 


1 833 


i8 9 3 


1932 


2012 


20T2 


60 


728 


2l3l 


2191 


225l 


23lO 


2370 


243o 


2489 


2349 


2608 


2668 


60 


729 


2728 


2787 


2S47 


2906 


2966 


3o25 


3o83 


3i44 


32o4 


3263 


60 


73o 


863323 


3382 


3442 


35oi 


356i 


3620 


368o 


3739 


3799 


3858 


5 9 


7 3i 


3917 


3977 


4o36 


4096 


4i55 


4214 


4274 


4333 


4392 


4452 


59 


732 


45 11 


4570 


463o 


4689 


4748 


4808 


4867 


4926 


4985 


5o45 


5 9 


7 33 


5io4 


5i63 


5222 


5282 


5341 


5400 


5459 


55 19 


5578 


5637 


5 9 


7 34 


5696 


5755 


58i4 


53 7 4 


5 ? 33 


5 99 2 


6o5i 


6110 


6169 


6228 


5 9 


735 


6287 


6346 


64o5 


6465 


6324 


6583 


6642 


6701 


6760 


6819 


5 9 


7 36 


6878 


6937 


6996 


7o55 


7114 


7173 


7232 


7291 


735o 


7409 


5 9 


737 


7467 


7526 


7385 


7644 


7703 


7762 


7821 


7880 


7939 


7998 


5 9 


7 38 


8o56 


8n5 


8174 


8233 


8292 


835o 


8409 


8468 


8327 


8586 


5 9 


7 3 9 


,8644 


8 7 o3 


8762 


8821 


8879 


8 9 38 


8997 


9036 


9n4 


9173 


5 9 


740 


869232 


9290 


9349 


9408 


9466 


9525 


9584 


9642 


9701 


9760 


5 9 


74i 


9818 


9877 


9935 


9994 


••53 


•111 


•170 


•228 


•287 


•345 


5 9 


742 


870404 


0462 


0321 


0379 


o638 


0696 


0755 


o8i3 


0872 


0930 


58 


743 


0989 


1047 


I I06 


1164 


1223 


1281 


i339 


i3 9 8 


1456 


I3l5 


58 


744 


i573 


i63i 


169O 


1748 


1806 


1 865 


i 9 23 


1981 


2040 


2098 
2681 


58 


745 


2 1 56 


22l5 


2273 


233 1 


238 9 


2448 


25o6 . 


2564 


2622 


58 


746 


2739 


2 797 


2855 


2913 


2972 


3o3o 


3o88 


3i46 


32o4 


3262 


58 


747 


332i 


33 79 


3437 


34g5 


3553 


36 1 1 


3669 


2727 , 


3785 


3844 


58 


748 


3902 


3960 


4018 


4076 


4i34 


4192 


425o 


4 3 08 


4366 


4424 


58 


749 


4482 


434o 


4598 


4656 


47U 


4772 


483o 1 


4888 


4945 


5oo3 


58 


l5o 


875061 


5i 19 


5i 77 


5235 


5293 


535i 


5409 


5466 


5524 


5582 


58 


7 5: 


564o! 


56 9 8 


5736 


58i3 


58 7 i 


5929 • 


5 9 S 7 


6o45 


6102 


6160 


58 


702 


6218, 


6276 


6333 


6391 


6449 


65o7 


6564 


6622 


6680 


6737 


58 


753 


6795, 


6853 


6910 


6968 


7026 


7083 


7141 


7199 


7256 


73 14 


58 


754 


7371! 


7429 


7487 


7344 


7602 


7639 


77n 


7774 


7832 


7889 


58 


755 


79471 8004 


8062 


8119 


8177 


8234 


8292 


8349 


8407 


8464 


57 


756 


8522 i 8579 


8637 


8694 


8732 


8809 


8866 


8924 


8981 


9039 


57 


7 5 7 


9096 1 9153 


92 1 1 


9268 


g323 


9383 


9440 


9497 


9555 


961? 


2 7 


758 


9669 9726 


9784 


9841 


9898 


9936 


••i3 


••70 


•127 


•i85 


57 


759 


880242 0299 


o356 


04i3 


0471 


o528 


o585 


0642 


0699 | 

8 


0736 


57 





1 


2 


3 


4 


5 


6 


^ ! 


9 


D. 



TABLE OF LOGARITHMS FROM 1 TO 10,000. 13 




U 



A TABLE OF liOGAEITHMS FEOM 1 TO 10,000. 



- 
N. 

820 





1 


! » 


3 


4 i 5 

1 


i 

6 


7 


8 


9 


D. 

53 


9j38i 


A 3S6 7 


3920 


3973 


4026 


! 4079 


4i32 


4184 


4237 


4290 


821 


4343 j 43p6 


4449 


4302 


4555 


4608 


4660 


47i3 


.4766 


4839 


53 


822 


4872 4925 


4977 


5o3o 


5o83 


5336 


5189 


5241 


5294 


5347 


53 


823 


54oo! 5453 


55o5 


5558 


56n 


5664 


5736 


5769 


5822 


5S75 


53 


824 


5927 5980 


6o33 


6o85 


6i38 


6591 


6243 


6296 


6349 


6401 


53 


825 


6454,' 6507 


6559 


6612 


6664 


6737 


6770 


6822 


63 7 5 


6927 


53 


826 


698c 


> 7033 


7085 


7i38 


7390 


7243 


7295 


7348 


7400 


7453 


53 


827 


75o( 


> 7558 


761 1 


7663 


7736 


7768 


7820 


7873 


7925 


7978 


52 


828 


8o3c 


:; 8o83 


8s35 


8188 


8240 


8293 


8345 


8397 


8430 


85o2 


5a 


829 


855: 


>: 8607 


8639 


8732 


8764 


8816 


8869 


8923 


8 97 3 


9026 


52 


83o 


91907^ 


>| 9i3o 


j 9183 


9235 


9287 


9340 


9 3 9 2 


9444 


9496 


9549 


52 


83 1 


960J 


1 9653 


; 9706 


97 58 


9810 


9S62 


9914 


9967 


•s.39 


e© 7 , 


52 


832 


92012.] 


0176 | 0228 


0280 


o332 


o384 


0436 


0489 


o543 


0093 


52 


833 


064: 


0697 


0749 


080 1 


o853 


j 0906 


0958 


1030 


1062 


II 14 


52 


834 


1166 


1218 


1270 


l322 


i374 


1426 


U78 


?53o 


i582 


1 634 


52 


835 


1686 


1738 


1790 


1842 


1894 


1946 


3998 


,2o5o 


2302 


2354 


5* 


836 


2206 


2258 


23 10 


2362 


2434 


2466 


2538 


2570 


2622 


2674 


52 


837 


,723 


2777 


?? 2 9 


2881 


2933 


2 9 85 


3o37 


3089 


3 140 


3392 


52 


838 


3244 


! 3296 


3348 


3399 


345i 


35o3 


3555 


3607 


3658 


3710 


52 


83 9 


3762 


j 38i4 


3865 


3917 


3969 


4021 


4072 


4124 


4176 


4228 


52 1 


840 


924279 


433 1 


4383 


4434 


44S6 


4538 


4589 


4641 


4693 


4744 


52 j 


841 


479k 


: 4848 


4899 


4951 


5oo3 


5o54 


5306 


5so7 


5209 
5723 


526i 


52! 


842 


53i2 


5364 


54i D 


5407 


55 s8 


5570 


562i 


56 7 3 


5776 


52 


843 


D828 


58 79 


5931 


5982 


6o34 


608 5 


6137 


6388 


6240 


6291 


5i 


844 


6342 


63* 


6445 


6497 


6348 


6600 


665j 


6702 


6754 


680 5 


'n 


840 


6S57 


6908 


6939 


703 1 


7062 


7* 1-4 


7t65 


7236 


7268 


73i 9 


31 


846 


7370 


7422 


7473 


7524 


7576 


7627 


7678 


773o 


77S1 


7S3 2 


5i 


847 


7883 


79 35 


7986 


8037 


8088 


8340 


9391 


8242 


8293 


8345 


5i 


848 


8396 


8447 


8498 


8549 


860 1 


8652 


8703 


8754 


88o5 


885 7 


5i 


849 


8908 


8959 


9010 


9061 


9112 


9i63 


92i5 


9266 


93i 7 


9368 


5i 


85o 


929419 


9470 


9521 


9 5 7 2 


9623 


9674 


9725 


9776 


9827 


9879 


5i 


85i 


9930 


9981 


••32 


• e 83 


•i34 


•i85 


•236 


•287 


•338 


•38 9 


5i 


852 


930440 


0491 


o542 


0592 


0643 


0694 


0745 


0796 


0847 


0898 


5i 


853 


0949 


1000 


io5i 


1102 


11 53 


1204 


1254 


i3o5 


i336 


1407 


5* 


854 


1458 


1 509 


i56o 


1 610 


1661 


1732 


1763 


1834 


3 865 


1935 


5- 


1 855 


1966 


2057 


2068 


2118 


2369 


2220 


2271 


2322 


2372 


2423 


5i 


1 8o6 


2474 


2024 


2575 


2626 


2677 


2727 


2778 


2829 


2879 


2930 


5i 


1 827 


2981 


3o3i 


3o82 


3333 


3i83 


3234 


3285 


3335 


3386 


343 7 


5i 


858 


3487 


3538 


353 9 


3639 


3690 


3 7 4o 


3791 


3843 


38 9 2 


3943 


5i 


8D9 


3993 


4o44 


4094 


414 5 


4190 


4246 


4296 


4347 


4397 


4448 


5i 


860 


934498 


4549 


4599 


465o 


4700 


475i 


4801 


4852 


4902 


4953 


5c 


861 


5oo3 


5o54 


5 1 04 


5i54 


52o5 


5255 


53o6 


5356 


5406 


5457 


5o 


8^2 


55p7 


5558 


56o8 


5658 


5709 


5739 


58o 9 


586o 


5910 


5960 


5o 


8o3 


601 1 


6061 


6111 


6162 


6232 


6262 


63i3 


6363 


64i3 


6463 


5oi 


864 

1 r\ * r 


65i4 


6564 


6614 


6665 


6 7 i5 


6765 


68i5 


6865 


6916 


6966 


5o 


86d 


7016 


7066 


7117 


7167 


7217 


7267 


7317 


7 36 7 


74i8 


7468 


5o 


866 


7018 


7568 


7618 


7668 


7718 


7769 


7819 


7869 


7939 


7969 


5o 


867 


8019 


8069 


S119 1 


8169 


8219 


8269 


832o 


8370 


8420 


8470 


5o 


868 


852o 


8D70 


8620 


8670 


8720 


8770 


8820 


8870 


8920 


8970 


5o 


869 


9020 


9070 


9120 


9170 


9220 


9270 


9320 


9369 


9419 


9469 


5o 


£70 i< 


?3 9 5i9' 


9569 


9619 


9669 


9719 


9769 


9819 


9869 


9918 


0968 


5o 


871 


540018 


0068 


0118 


0168 


0218 


0267 


o3i7 


o367 


0437 


0467 


5o 


872 


o5i6 


o566 


0616 


0666 


0716 


0765 


o8i5 


o865 


0933 


0964 


5o 


873 


1014J 


1064 


1 1 14 


n63 


12l3 


1263 


i3i3 


i362 


1412 


1462 


5o 


874 


i5i 1 


i56i 


361 1 


1660 


1710 


1760 


1809 


1859 


1909 


1958 


5o 


875 


2008 


2o58 


2107 


2157 


2207 


2256 


23o6 


2355 


24o5 


2455 


5o) 


876 


25o4 


2554 


26o3 


2653 


2702 


2752 


2801 


285i 


2901 


2900 


5o 


877 


3ooo 


3 049 


3099 


3i48 


3398 


3247 


3297 


3346 


3396 


3445 


49 


878 


3495 


3544 


3593 


3643 


3692 


3742 


3791 


384i 


3S90 


3939 


4o 1 


879 


39891 


4o38 


4088 


4i3 7 


4386 


4236 


.*285 


4335 


4384 


4433 


49 


K. 





1 


2 


3 | 


4 


5 


6 


7 

1 


8 


I 

9 i 



A TABLE OF LOGARITHMS FROM 1 TO 10,000. 



15 



88o 
83 1 
882 
883 
884 
885 
886 
887 
888 
889 

890 
891 
892 
893 
894 
89.5 
896 

897 
898 

899 

900 
901 
902 
903 
904 
905 
906 
907 
908 
909 



944433. 

4976! 

596 1 
6432! 
6943' 6992 



4532 

5023 

55i8 
6010 
65oi 



7434, 
7924, 
84i3, 
8902 

949390 

,9^73 
95o365 
o35i 
i338 
1323 
23o8 
2792 
3276 
3760 

954243 
4725 
D207 
56b3 
6168 
6649 
7120 
7607 
boob 
8564 



9^9041 
9D18 
9995 

90047 1 
0946 
1421 

2369 
2343 
33i6 



910 
911 
912 
9i3 
914 
g 1 5 
916 

9'7 
918 

9 r 9 

920 963783 

92 1 
922 
923 



924 
925 
926 

9^7 
928 

929 

^lo 
9 !i 
9 32 
9 33 

934 

935 
9 36 

9 3 7 
9 33 

9 3 9 

N. 



4260 
473i 

5202 
5672 
6142 
66l I i 
7080' 

7548 

8016 



7433 

7973 
8462 
8951 

9439 
9926 
o4U 

0900 
1386 
1872 
2356 
2841 
3325 
33o3 

4291 

4773 
5255 
5 7 36 
6216 
6697 
7176 
7655 

8i34 
8612 

90S9 
cpbb 
••42 
o5i8 
0994 
1469 
194J 

2417 

23qo 

3363 

3835 
4307 
4773 
5249 
5719 
6i8g 
6658 
7127 
7 5 9 3 
8062 



853o 
8996 

9463 
9028 



968483 

8950: 

0416 1 

9882 
97o347j 6^o3 

33 1 21 0838 

1276 i32J 

1740 1786 

2203 2249 
2666 2712 



458 1 
5074 
5567 
6039 
655i 
7041 
7532 
8022 
85 1 1 
8999 

9488 
997 5 
0462 
0949 
1435 
1920 
24o5 
2889 
337i 
3856 

4339 
4821 
53o3 

5 7 84 
6265 
6745 
7224 

77°3 
8181 
8659 

9 t3 7 

9614 
••90 
o566 
1041 

I3l6 

1990 

2464 
2 9 3 7 
34io 

3882 

4354 
4825 

5296 
5766 
6236 
6705 
7173 
7642 
8109 

85 7 6 
9043 

9^°9 
9973 

0440 
090 \ 
1 J 69 
i832 

2293 

2 7 33 



7 



8 



463 1 
5i24 
56i6 
6108 
6600 
7090 
753i 
8070 
8360 
9048 

9536 
••24 
o5u 
0997 
1483 
1969 
2453 
2 9 38 
342,1 
3905 

438 7 
4869 
535i 
5^32 
63 1 3 
6 79 3 
7272 
7731 
8229 
8707 

9183 
9661 
•i38 
061 3 

io3 9 
1 56 \ 

20 33 

2 31 I 
2983 
3457 

3929 
4401 
4372 
5343 

53i3 
6233 
6732 
7220 
7688 
8i56 

8623 
9090 
9556 
••21 
o i86 
0931 

I 4 I 3 
1S79 
2342 
2804 



I " 



4680 
5l73 

5665 
6157 
6649 
7140 
763o 
8119 
8609 
9097 

9 5S5 
•• 7 3 
o56o 
1046 
i532 
2017 

25o2 

2986 
3470 
3953 

4435 
4918 

5399 

588o 
636i 

6340 
7320 

7799 
8277 
8 7 55 

9232 
9709 
•i8d 
066 1 
1 136 
161 1 
2o85 
2539 
3o32 
35o4 

3977 
4448 

4919 
5390 

586o 

6329 
6799 

7267 

82o3 
8670 

9 I 33 

9602 
••68 
o533 
0997 
1461 
1923 
2388 
285i 



4729 

5222 
57l5 
6207 
6698 
7189 
7679 
8168 

8657 
9146 

9634 
•121 
0608 

1095 
i5 v :o 
2066 
255o 
3o34 
35i8 
4001 

4434 
4966 

5447 
5928 

6409 
6383 
7368 

7847 
8325 
83o3 

9280 

97 ""'7 
•233 

0709 

n84 

1638 

2l32 

2606 
3079 

3552 



4024 

449 5 

4966 

5437 
6907 
63 7 6 
6345 
73 1 4 

8249 
8716 
9 is3 
9649 
• 1 1 4 
0579 

1 044 

1 5o8 

1971 

2434 

2S97 



4779 
5272 

5764 ; 

6256 

6747 
7233 
7728 
8217 
8706 
9193 

9 683 
•170 
0657 
1 1 43 
1629 
2114 
2299 
3o83 
3566 
4049 

4532 
5oi4 
5493 

5976 

6457 
6 9 36 

74i6 

7V 
8373 

8S5o 

9328 

9804 
•280 
0736 
1 23 1 

1706 
2180 
2653 
3i26 
3599 

4071 
4342 
5oi3 
5484 
5 9 34 
6423 
6892 
736i 
7829 
8296 

8763 

9229 
9693 
•161 
0626 
1090 
1 5 5 1 

20l3 

2481 
2943 



4828 

532i 
58i3 
63o5 
6796 

7287 

7777 
8266 

8755 

9244 

973 1 
•219 
0706 
1192 
1677 
2i63 
2647 
3i3i 
36i5 
4098 

458o 
5o62 
5543 
6024 
65o5 
6984 
7464 
7912 
8421 
8898 

9 3 7 5 

o332 

•328 
0804 

12-9 

17 )3 
2227 
2701 
3174 
3646 



4i 18 

4390 
5o6i 
553 1 
6001 
6470 
6939 
74o3 
7875 
8343 

8810 
9276 
9712 
•207 
0672 
1 1 37 
1 60 1 
2064 

2327 
2 9 3 9 



4877 
5370 

5862 
6354 
6845 
7336 
7826 
83i5 
8804 
9292 

978o 
•267 
0754 
1240 
1726 
2211 
2696 
3i8o 
3663 
4146 

4628 
5i 10 
5592 
6072 
6553 
7032 
7312 
7990 
8408 
8946 

9423 
9900 
•376 
08 5 1 
1 326 
1 801 
2273 
2748 

32 2 1 
3693 

4 1 65 
4637 
5io3 
5573 
6048 
6317 
6986 
7454 
7922 
8890 

8856 
932 \ 
9789 

•2 5 i 

0719 
118} 

1647 

2110 

3o35 
8 



4927 
5419 
5912 
640 3 
6894 
73^3 
7 3 7 5 
8364 
8853 
934i 

9829 

•3i6 
o8o3 

1289 
1773 
2260 

2744 

3223 

3711 

4194 

4677 

5i58 
6640 
6120 
6601 
7080 
7559 
8o38 
85i6 
8994 

947 1 
9947 
•423 

0899 
i374 
1848 

2322 
2795 

3268 

3741 

4212 

4684 

5i55 

5623 

6095 
6564 
7 o33 
75oi 
7969 
8436 

8903 
9 36q 
9835 
•3oc 
0763 
1229 
i6o3 
2 1 37 
2619 
to8a 



D. 

49 
49 
49 
49 
*9 
49 
49 
49 
49 
49 

49 
49 
49 
49 
49 
48 

48 
48 
48 

48 

43 
48 
43 
48 
43 
48 
48 
48 
48 
48 

48 
48 
48 
48 
47 
47 
47 
47 
47 
47 

47 
47 
4" 
47 
47 
47 
47 
47 
47 
4: 

47 
47 
47 
47 
46 
46 
46 

46 

46 
46 



D. 



26 



16 



A TABLE OF LOGARITHMS FROM 1 TO 10,000. 



N. 





1 


l 

1 ' 


3 


4 


5 


6 


7 


8 


9 


IX 


94o 


973128 


1 3i 7 4 


3220 


3266 


33i3 


335 9 


34o5 


345i 


3497 


3543 


46 


941 


3390 


3636 


3682 


3728 


3774 


3820 


3866 


3913 


3959 


4oo5 


46 


942 


4o5i 


4097 


4i43 


4189 


4235 


4281 


4327 


4374 


4420 


4466 


46 


943 


4312 


4558 


4604 


465o 


4696 


4742 


4788 


4834 


4880 


4926 


46 


944 


4972 


5oi« 


5o64 


5no 


5i56 


5202 


5248 


5294 


534o 


5386 


46 


945 


54 J 2 


5478 


5524 


5570 


56i6 


5662 


5707 


5733 


5 799 


5845 


At 


946 


58 9 i 


5 9 3 7 


3983 


6029 


6075 


6121 


6167 


6212 


6258 


63o4 


46 


947 


635o 


63 9 6 


6442 


6488 


6533 


6579 


6625 


6671 


6717 


6 7 63 


46 


948 


680S 


6854 


6900 


6946 


6992 


7037 


7083 


7129 


7175 


7220 


4*j 


949 


7266 


7312 


7 358 


74o3 


7449 


7495 


7541 


7586 


7632 


7678 


46 


930 


977724 


7769 


7815 


7861 


7906 


79 52 


7998 


8o43 


8089 


8i35 


46 


931 


8181 


8226 


8272 


8317 


8363 


8409 


8454 


85oo 


8546 


85 9 i 


46 


932 


8637 


8683 


8728 


8774 


8819 


8865 


8911 


8 9 56 


9002 


9047 


46 


933 


9093 


9i38 


9184 


9230 


9275 


9321 


9 366 


94i2 


9457 


95o3 


46 


954 


9348; 9394 


9639 


9 685 


9730 


9776 


9821 


9867 


9912 


99 58 


46 


955 


980003 


0049 


0094 


0140 


oi85 


023l 


0276 


0322 


0367 


0412 


45 


936 


0438 


o5o3 


0549 


0594 


0640 


o685 


0730 


O776 


0821 


0867 


45 


937 


0912 


0957 


ioo3 


1048 


1093 


1 1 39 


1 184 


1229 


1275 


l320 


45 


958 


1366 


1411 


1456 


i5oi 


1547 


l5g2 


1637 


i683 


1728 


1773 


45 


969 


1819 


1864 


1909 


1954 


2000 


2045 


2090 


2i35 


2181 


2226 


45 


960 


982271 


23i6 


2362 


2407 


2432 


2497 


2543 


2588 


2633 


2678 


45 


961 


2723 


2769 


2814 


2859 


2904 


2949 


2994 


3 040 


3o85 


3i3o 


45 


962 


3i75 


3220 


3265 


33io 


3356 


34oi 


3446 


3491 


3536 


358i 


45 


963 


3626 


36 7 i 


3716 


3762 


3807 


3852 


38 97 


3o42 


3987 


4o32 


45 


964 


4077 


4122 


4167 


4212 


4257 


43o2 


4347 


4J92 


4437 


4482 


45 


963 


4327 


4572 


4617 


4662 


4707 


4752 


4797 


4842 


4887 


4932 


45 


966 


4977 


5022 


5067 


5ll2 


5i 57 


5202 


5247 


5292 


5337 


5382 


45 


967 


5426 


5471 


55i6 


556i 


56o6 


565i 


56 9 6 


5741 


5 7 86 


583o 


45 


968 


58 7 5 


5920 


5965 


6010 


6o55 


6100 


6144 


6189 


6234 


6279 


45 


969 


6324 


6369 


64i3 


6458 


65o3 


6548 


65 9 3 


6637 


6682 


6727 


45 


970 


986772 


6817 


6861 


6906 


6 9 5i 


6996 


7040 


7o85 


7i3o 


7173 


45 


97i 


7219 


7264 


7 3o 9 


7 353 


73 9 8 


7443 


7488 


7 532 


7577 


7622 


45 


972 


7666 


7711 


7756 


7800 


7845 


7890 


7934 


7979 


8024 


806S 


45 


973 


81 13 


8107 


8202 


8247 


8291 


8336 


838i 


8423 


8470 


85i4 


45 


974 


855 9 


8604 


8648 


86 9 3 


8737 


8782 


8826 


8871 


8916 


8960 


45 


973 


9003 


9049 


9094 


9 i33 


9i83 


9227 


9272 


93i6 


9J61 


9403 


45 


976 


g45o 


94Q4 


9539 


9 583 


9628 


9672 


9717 


9761 


9806 


9.85o 


44 


977 


9893 


9939 


9983 


o» 2 8 


•0^2 


•117 


•161 


®206 


°25o 


•294 


44 


•978 


9903J9 


o383 


0428 


0472 


o5i6 


o56i 


o6o5 


o65o 


0694 


o 7 33 


44 


979 


0783 


0827 


0871 


0916 


0960 


1004 


1049 


1093 


1137 


1182 


44 


980 


991226 


1270 


i3i5 


i359 


i4o3 


1448 


1492 


1 536 


i58o 


i625 


44 


981 


1669 


I7i3 


i 7 58 


1802 


1846 


1890 


1935 


1979 


2023 


2067 


44 


982 


21 1 1 


2 1 56 


2200 


2244 


22S8 


2333 


2377 


2421 


2465 


2309 


44 


9 83 


2554 


2598 


2642 


2686 


2730 


2774 


2819 


2863 


2907 


2g5i 


44 


984 


2995 


3o39 


3o83 


3127 


3172 


3 2 i6 


3260 


33o4 


3348 


3392 


44 


9 85 


3436 


3480 


3524 


3568 


36 1 3 


3657 


3701 


3745 


3789 


3833 


44 


986 


38 77 


3921 


3965 


4009 


4o53 


4097 


4UT 


4i85 


4229 


4273 


44 


987 


43i7 


436 1 


44o5 


4449 


449 3 


4537 


458i 


4625 


4069 


47'3 


44 


988 


4757 


4801 


4845 


4889 


4933 


4977 


5021 


5o65 


5 1 08 


5i52 


44 


989 


5i 9 6 


0240 


5284 


5328 


53 7 2 


54i6 


5460 


55o4 


5547 


5591 


44 


990 


995635 


56 79 


5723 


5767 


58n 


5854 


58 9 8 


5942 


5 9 86 


6o3o 


44 


991 


6074 


61 17 


6161 


62o5 


6249 


6293 


6337 


633o 


6424 


6468 


44 


99 2 


65i2 


6555 


6599 


6643 


6687 


6 7 3i 


6774 


6818 


6862 


6906 


44 


993 


6949 


699.3 


7037 


7080 


7124 


7168 


7212 


7255 


7299 


7343 


44 


994 


7 386 


743o 


7474 


7517 


756i 


76o5 


7648 


7692 


77 36 


7779 


44 


993 


7823 


7867 


79'° 


7954 


7998 


8041 


8o85 


8129 


8172 


8216 


44 


996 


8239 


83 o3 


8347 


83 9 o 


8434 


8477 


852i 


8564 


8608 


8652 


44 


997 


86 9 5 


8 7 3 9 


8782 


8826 


8869 


8913 


8 9 56 


9000 


9043 


9087 


44 


998 


9 1 3 1 


9'74 


9218 


9261 


93o5 


9348 


9392 


9435 


9479 


9322 


44 


999 
N. 


9365 


9609 


9632 


9696 


9739 


9783 


9826 


9870 


99 i3 


99 5 7 


43 





1 


3 


3 


4 


5 


6 


7 


8 


9 


D. 



A TABLE 



OF 



LOGARITHMIC 
SINES AND TANGENTS 



FOR EVERY 



DEGREE AND MINUTE 
OF THE QUADRANT. 



Remark. The minutes in the left-hand column of each 
page, increasing downwards, belong to the degrees at the 
top ; and those increasing upwards, in the right-hand column, 
belong to the degrees below. 



18 



(0 DEGEEES.) A TABLE OF LOGARITHMIC 



ML 




Sine 


D. 


Cosine 


D. 


Tang. 


D. 


Cotang. 


60 


o- 000000 




10-000000 




0-000000 




Infinite. 


i 


6.463726 


5017.17 


00000c 


• 00 


6-463726 


5017-17 


13.536274 


5 9 


2 


764736 


2934-85 


000000 


• 00 


764756 


2 9 34-83 


235244 


58 


3 


940847 


20S2'3l 


000000 


• 00 


940847 


2082 -3i 


059163 


57 


4 


7 '060786 


i6i5-i7 


000000 


■ 00 


7-o65 7 86 


i6i5-i7 


12 -934214 


56 


5 


162696 


i3i 9 -68 


000000 


• 00 


162696 


i3iq.6o 


837304 


55 


6 


241877 


1115.75 


9-999999 


•01 


241878 


1113-78 


758122 


54 


7 


308824 


966-53 


999999 


•01 


3o8825 


096-53 


691175 


5i 


8 


3668i6 


852-54 


999999 


•01 


366817 


852-54 


633i83 


D2 


9 


j 417968 


762-63 


999999 


•01 


417970 


762-63 


582o3o 


5i 


IO 


463725 


689-88 


999998 


•01 


463727 


689-88 


536273 


5o 


ii 


7-5o5n8 


629.81 


9-999998 


•01 


7'5o5i2o 


629-81 


12-494880 


4Q 


12 


542906 


579.36 


999997 


•01 


542909 


579-33 


4^7091 


4» 


i3 


577668 


536-41 


999997 


•01 


577672 


536-42 


422328 


47 


U 


609853 


499-38 


999996 


•01 


609857 


499 -3o 


390143 


46 


if 


63g8i6 


467-14 


999996 


•01 


639820 


467.15 


36oi8o 


45 


16 


667845 


438-8i 


999995 


•01 


667849 


438-82 


332i5i 


AA 


17 


694173 


413.72 


999995 


•01 


694179 


4i3-73 


3o582i 


43 


18 


718997 


391-35 


999994 


•01 


719004 


391.36 


280997 


42 


19 


742477 


371-27 


999993 


•01 


742484 


371-28 


257316 


4i 


20 


764754 


353-i5 


999993 


•01 


764761 


35i-36 


236239 


4o 


21 


7 -785 9 43 


336-72 


9.999992 


•01 


7.785951 


336- 7 3 


12 .214049 


3 9 


22 


806146 


321-75 


999991 


•01 


8o6i55 


321-76 


193843 


38 


23 


82545i 


3o8-o5 


999990 


•01 


825460 


3o8-o6 


174540 


37 


24 


843934 


295-47 
2S3-88 


999989 


•02 


843944 


295.49 


i56o56 


36 


25 


861662 


999988 


•02 


861674 


283.90 


138326 


35 


26 


878695 


273-17 

263-23 


999988 


•02 


878708 


273-18 


121292 


34 


27 


895085 


9999 8 7 
999986 


•02 


895099 


263-25 


1 0490 1 


33 


28 


910879 


253-99 


•02 


910894 


254-oi 


080106 


32 


29 


926119 


245-38 


999985 


•02 


926134 


245-4o 


07^866 


3i 


3o 


940842 


237.33 


999983 


•02 


94o858 


237.35 


069142 


3o 


3-i 


7-955082 


229-80 


9-999982 


•02 


7.955100 


229-81 


1 2 • 044900 


29 


32 


968870 


222-73 


999981 


•02 


968880 


222-75 


o3im 


28 


33 


982233 


216-08 


999980 


•02 


982253 


2l6-I0 


017747 


27 


34 


995198 

8-007787 


209-81 


999979 


•02 


995219 


2oo-83 


004781 


26 


35 


203-90 


999977 


•02 


8-007809 


203-92 


11-992191 


25 


36 


020021 


198.31 


999976 


•02 


020043 


198.33 


979935 


24 


u 


031919 


193-02 


999975 


•02 


o3io45 


193-05 


968055 


23 


043301 


188-01 


999973 


•02 


043627 


i88-o3 


956473 


22 


39 


034781 


183-25 


999972 


•02 


054809 


i83-2 7 


945191 


21 


40 


065776 


178-72 


999971 


•02 


o658o6 


178-74 


934194 


20 


41 


8-076500 


I74-4I 


9-999969 


•02 


8-07653i 


174-44 


11-923469 


IG 


42 


086965 


170-3 1 


999968 


• 02 


086997 


170-34 


9i3oo3 


10 


43 


097183 


166. 3o 


999966 


•02 


097217 


166.42 


002783 


17 


44 


107167 


162-63 


999964 


•o3 


107202 


162-68 


892797 


16 


45 


1 16926 


159-08 


Q99963 


•o3 


1 i6o63 


159. 10 


883o37 


i5 


! 46 


1 2647 1 


1 53 -66 


999961 


•o3 


126310 


155-68 


873490 ■ 


14 


47 


i358io 


152-38 


999959 


•o3 


i3585i 


i52-4i 


864149 


i3 


48 


144953 


149-24 


999958 


•o3 


144996 


149-27 


855oo4 


12 


49 


153907 


146-22 


9999D& 


•o3 


153932 


146-27 


846048 


11 


5o 


162681 


143-33 


999954 


•o3 


162727 


U3-36 


837273 


10 


5i 


8-171280 


140 -54 


9-999952 


•o3 


8-171328 


140-57 


11-828672 


Q 


52 


179713 


137.86 


9 999 5o 


•o3 


179763 


137-90 


820237 


8 


53 


187985 


135.29 


999948 


•o3 


i88o36 


i35-32 


81 1964 

8o3844 


7 


54 


196102 


i32-8o 


999946 


•o3 


196156 


132-84 


6 


55 

56 


204070 
21 1895 


i3o-4i 
128-10 


999944 
999942 


•o3 
•04 


204126 
211953 


i3o-44 
128-14 


795874 
788047 


5 

4 


5 7 


219581 


125-87 


999940 


•04 


2 1 964 1 


125-90 


78o35o 


3 


58 


227134 


123-72 


999938 


•04 


227195 


123-76 


772803 


2 


5 9 


23455] 


121-64 


999936 


•04 


234621 


121-68 


763379 


1 


60 


24i853 


119-63 


999934 


•04 


241921 


1 19-67 


758079 



M. 


Cosine 


D. 


Sine 


1 
1 


Cotang, 


D. 


Tang. 



(89 DEGREES.) 



SIXES AND TANGENTS. (1 DEGREE.) 



19 



[~M. 



Sine 



o 
i 

2 

3 

4 , 

6 I 

I 

V 

9 

10 



II 

12 

i3 

14 
i5 
16 

i 
i 

l 9 

20 

21 

22 

23 

24 

25 

26 

2 

2 

3o 

3i 

32 

33 

34 

35 

36 

3 

3 

3 9 

4o 

4i 
42 
43 
44 
45 
4b 

47 
4* 

5o 
5i 

52 

53 

54 

55 

56 

5 

5 

5q 

6o 



D. 



■24i855 
249o33 
206094 
263042 
269881 
276614 
283243 
289773 
296207 
302046 
308794 

• 3 1 4904 
321027 
327016 
332924 
338753 
3445o4 
35oi8i 
355783 
36i3i5 
366777 

•372171 

377499 
382762 

387962 

393101 

398179 

403199 

408161 

4i3o68 

417919 

422717 
427462 
432 1 56 
4368oo 
44i394 
445941 
45o44o 
454893 
459301 
463665 

8-467985 
472263 

476498 
480693 
484848 
48s 9 63 
493o4o 
497078 
5oio8o 
5o5o45 

8 .508074 
512067 
516726 
52o55i 
5243 
52H102 
531828 
535523 
539186 
542819 



Cosine 



8^ 



Cosine 



119-63 
117-68 
n5-8o 
n3- 9 8 

112-21 
II0-50 

io8-83 
107-21 
io5-65 
io4-i3 
102-66 

101-22 

90-82 
98-47 
97.14 
9 5-86 
04.60 
9 3 -38 
92-19 
91 -o3 
89-90 

88- 80 

87-72 
86-67 
85-64 
84-64 
83-66 
82-71 

81-77 
80 -80 
79.96 

79.09 
78-23 
77-40 
76-57 

75-77 

74-99 
74-22 

73.46 
72-73 
72-00 

71-29 
70-60 
69-91 
69-24 
68.59 
67-94 
67.31 
66-69 
66- 08 
65-48 

64-89 
64-3i 
63- 7 5 
63-i 9 
62-64 
62-11 
6i-58 
61 -06 
60 -55 
60-04 



D. 



Tan?. 



9.999934 
999932 
999929 
999927 
999920 
999922 
999920 
999918 

9999*5 
999913 
999910 

9.999907 
999905 
999902 
999899 
999897 
999894 
999891 
999888 
999885 
999882 

9-990879 
999876 
999873 
999870 
999867 
999864 
999861 
999858 
999854 
999851 

9.999848 
999844 
999841 
99983 8 
999834 
999831 
999821 
999823 
999820 
999816 

9.999812 
999809 
99980 D 
99980 1 

990797 
999793 
999700 
999786 
999782 
999778 

9-999774 
999769 
999763 
999761 
999757 
999753 
999748 
999744 
999740 
999733 



D. 



• 04 

.04 
•04 
•04 

• 04 

• 04 
•04 

• 04 
.04 
•04 
.04 

•04 
•04 
•Oi 

• o5 
•o5 

• o5 

• o5 

• o5 

• o5 

• o5 

•o5 

• o5 

• o5 

• o5 

• o5 

• c5 

• o5 

• o5 

• o5 

• 06 

.06 

• 06 

• 06 
.06 
.06 

• 06 

• 06 

• 06 
.06 

• 06 

• 06 

• 06 
.06 
.06 

•07 
.07 

• 07 
•07 
•07 
•07 

• 07 

• 07 
•07 

•07 

• 07 

•07 
■07 

•07 
•07 
•07 



D. 



8.241921 
249102 
256i65 
263n5 
269936 
276691 
283323 
289836 
296292 
3o2634 
3oS884 

8-3i5o46 

32II22 
327114 

333025 
338856 
344610 
330289 
3558o3 
36i43o 
3668 9 5 

8.372292 
377622 
382889 
388oo2 
393234 
3 9 83i5 
4o3338 
4o83o4 
4i32i3 
418068 

8-422869 
427618 
4323i5 
436o62 
44i36o 
446 1 1 o 
45o6i3 
453070 
45o48i 
46J849 

8.468172 
472454 
476693 
480892 
485o3o 
489170 

4c;J25o 

497293 
501298 
503267 

8.5oq2go 
5i3o;3 
516961 
520790 
524586 
523349 
532o8o 
533779 

539447 
543o84 



Cot&ng. 



Sine 



Cotang. 



119-67 

117-72 
1 1 5 • 84 
114-02 

112-25 

no-54 
108-87 
107-26 
105.70 
104-18 
102-70 

101 .26 

99.87 
98.51 
97.19 
95.90 
9 4-65 
9 3-43 
92-24 
91.08 
89.95 

88-85 

87-77 
86-72 

85-70 

84-70 
83-71 
82-76 
81-82 
80-91 
80-02 

79.14 

78.30 
77-45 
76 -63 
75-83 
75. o5 
74-28 
73-52 
72.79 
72-06 

71-35 
70-66 
69.98 
6 9 -3i 
68-65 
68-01 
67-38 
66-76 
66* 1 5 
65-55 

64-96 
64-39 
63-82 
63-26 
62-72 
62-18 
6i-65 
6i-i3 
60-62 
60 • 12 



11 



I r 



758079 
750890 
743835 
736885 
730044 
723309 
716677 

710144 
703708 
697366 
691116 

■684954 
678878 
672886 
666975 

661144 
635390 

6497 ll 
644 io5 
638570 
633io5 

11-627708 
622378 
617111 
61 1908 
606766 
6oi685 
596662 
591696 
586787 
581932 

U.577131 
572382 
567685 
563o38 
558440 
553890 
549387 
544§3o 
540319 
536i5i 

11. 53i828 
627546 
5233o7 
51910S 
514950 
5io83o 
506750 
502707 
40870.! 
494733 

1 1 .490800 
48690a 
483o3g 
479210 
4754U 
4*7165] 
4^7920 
464221 
46o553 
436916 



60 

5? 
5 7 
56 
55 
54 
53 

52 

5i 
5o 

49 
48 

47 
46 
45 
44 
43 
42 
41 
40 

3 9 
38 

37 
36 

35 

34 
33 

32 

3i 
3o 

29 
28 

27 
26 

25 

24 

23 
22 
21 
20 

IO 

10 

17 
16 

i5 
14 
i3 

12 
1 1 
10 

Q 

6 

7 
6 
5 
4 
3 
2 
1 
o 



D. 



Tang 



(88 DEGREES.) 



20 


(2 


DEGREES.) A TABLE OF LOGARITHMIC! 




M. 




Sine 


D 




Cosine 


D. 


j Tang. 


D. 


Cotang. 




8.542819 


60 


.04 


9-999735 


• 07 


8- 543o84 


60-12 


ii'4569i6 


60 


i 


546422 


5 9 


-55 


999731 


.07 


546691 


59 


•62 


453309 


5q 


2 


54QQ95 


5 9 


• 06 


999726 


• 07 


55o268 


59 


•14 


449732 


58 


3 


', 553539 


58 


•58 


999722 


• 08 


5538i7 


58 


•66 


446i83 


n 


4 


557054 


58 


• 11 


999717 


.08 


55 7 336 


58 


•19 


442664 


5 


56o54o 


57 


• 65 


999713 


.08 


560828 


5 7 


•73 


439172 


55 


6 


563 999 


5 7 


.19 


999708 


.08 


564291 


57 


•27 


436709 


54 


7 


56743i 


56 


•74 


999704 


.08 


567727 


56 


•82 


432273 


53 


8 


5 7 o836 


56 


• 3o 


999699 


.08 


571137 


56 


•38 


428863 


52 


9 


574214 


55 


•87 


999604 


.08 


574520 


55 


•?5 


■ 425480 


5i 


10 


577566 


55 


■44 


999689 


.08 


577877 


55 


•02 


422123 


5o 


ii 


8-580892 


55 


• 02 


9-999685 


.08 


8-58i2o8 


55 


■ 10 


1 1 -418792 


40 


12 


584193 


54 


• 60 


999680 


.08 


5845i4 


54 


■68 


415486 


48 


•i3 


587469 


54 


J 9 


999675 


.08 


587795 


54 


•27 


412205 


47 


U 


590721 


53 


79 


999670 


.08 


591001 


53 


•87 


408949 


46 


i5 


593948 


53 


•3 9 


999665 


• 08 


5 9 4283 


53 


•47 


405717 


45 


16 


597152 


53 


00 


999660 


• 08 


597492 


53 


•08 


4o25o8 


44 


17 


6oo332 


52 


61 


999655 


.08 


600677 


52 


■70 


399323 


43 


18 


603489 


52 


23 


999650 


.08 


6o383o 


52 


•32 


3g6i6i 


42 


19 


606623 


5i 


86 


999645 


• o 9 


606978 


5i 


•Q4 


3o3o22 


4i 


20 


609734 


5i 


49 


999640 


.09 


610094 


5i 


• 58 


389906 


4o 


21 


8-612823 


5i 


12 


9-999635 


• 09 


8-613189 


5i 


21 


ii-3868n 


3 9 


22 


6i58oi 


5o 


76 


999629 


• o 9 


616262 


5o 


85 


383738 


38 


23 


618937 


5o 


4i 


999624 


-o 9 


619313 


5o 


5o 


380687 


n 


24 


621962 


5o 


06 


999619 


• o 9 


622343 


5o 


i5 


377657 


25 


624965 


49 


72 


999614 


• o 9 


625352 


49 


81 


374648 


35 


26 


627948 


49 


38 


099608 


• o 9 


628340 


49 


47 


371660 


34 


27 


63oou 


49 


04 


999603 


• 09 


63i3o8 


49 


i3 


3686 9 2 


33 


28 


633854 


48 


7 1 


999597 


.09 


634256 


48 


80 


365 7 44 


32 


J9 


636776 


48 


3 9 


999592 


.09 


637184 


48 


48 


362816 


3i 


3o 


639680 


48 


06 


999586 


.09 


640093 


48 


16 


359907 


3o 


3i 


8-642563 


47 


75 


9-999581 


.09 


8-642982 


47 


84 


11-357018 


29 


32 


645428 


47 


43 


99 9 5 7 5 


-o 9 


645853 


47 


53 


354U7 


28 


33 


648274 


47 


12 


999570 


.09 


648704 


47 


22 


351296 


27 


34 


65no2 


46- 


82 


909564 


.09 


65i537 


46 


9i 


348463 


26 


35 


65391 1 


46- 


52 


999558 


• 10 


654352 


46 


61 


345648 


25 


36 


656702 


46- 


22 


999553 


• 10 


657149 


46 


3i 


34285i 


24 


37 


659475 


45. 


92 


999547 


• 10 


659928 


46 


02 


340072 


23 


38 


66223o 


45- 


63 


999341 


• 10 


662689 


45 


73 


3373 1 1 


22 


3o 


664968 


45- 


35 


999535 


• 10 


665433 


45 


44 


334567 


21 


4o 


667689 


45- 


06 


999529 


• 10 


668160 


45 


26 


33i84o 


20 


4i 


8-67o3o3 


44. 


79 


9-999524 


• 10 


8-670870 


44 


88 


11-329130 


'A 


42 


678000 


44- 


5i 


999518 


• 10 


673563 


44 


61 


326437 


43 


675751 


44- 


24 


999512 


• 10 


676239 


44 


34 


323761 


\l 


44 


678405 


43- 


97 


999506 


• 10 


678900 


44 


17 


321 100 


45 


681043 


43- 


70 


999500 


• 10 


68iD44 


43 


80 


3 1 8456 


i5 


46 


683665 


43- 


44 


9994o3 


• 10 


684172 


43 


54 


3i5828 


14 


47 


686272 


43- 


18 


999487 


• 10 


686784 


43 


28 


3i32i6 


i3 


48 


688863 


42- 


92 


999481 


• 10 


68 9 38i 


43- 


o3 


310619 


13 


49 


691438 


42- 


67 


999475 


• 10 


691963 


42 


77 


3o8o37 n 


5o 


693998 


42- 


42 


999469 


• 10 


694029 ; 


42 


52 


300471 10 1 


5i 


8-696543 


42. 


17 


9-999463 


• 11 


8-697081 


42- 


28 


11-302919 


9 


52 


699073 


4i- 


9n 


999456 


• 11 


699617 


42- 


o3 


3oo383 


8 


53 


701589 


4i- 


68 


999450 


• 11 


702139 


4i- 


22 


297861 


7 


54 


704090 


4i- 


44 


999443 


• 11 


704646 


4i- 


55 


295354 


6 


55 


706577 


4i- 


21 


999437 


• 11 


707140 


4i- 


32 


292860 


5 


56 


709049 


40. 


97 


99943i 


• 11 


709618 


41- 


08 


2O0382 


4 


57 


711507 


40- 


74 


999424 


• 11 


712083 | 


4o- 


85 


287917 


3 


58 


713952 


40- 


5i 


999418 | 


• 11 


714534 


4o- 


62 


285465 2 


5g 


7i6383 


4o- 


20 


99941 1 : 


• 11 


716972 


4o- 


40 


283o28 


1 


6o 


718800 


4o-o6 


999404 


• 1 1 


719396 


40-17 


280604 



M. 


i 


Cosine 


D. 


Sine 




Cotang. 


D. 


Tang. . 






• 




(87 r 


iegr; 


RES.) 











SIXES AND TANCafifPTS. (3 DEGREES.) 



21 



M. 



Sine 



D. 



Conine L>. 



Tang. 



D. 






8--mS3oo 


I 


721204 


2 


723393 


3 


725972 


4 


— 337 


5 


73068S 


6 


733o27 


7 


735354 


8 


737667 


9 


739969 


10 


742239 


ii 


8-744536 


12 


746802 


1 3 


749033 


14 


731297 


13 


753528 


16 


755-4- 


17 


73" 


18 


760131 


19 


762337 


20 


76451 1 



21 

22 

23 

2 4 

23 

I -A 
3 

29 

3o 

3i 

! 32 
33 
34 
35 
36 
3 

39 
40 



41 
42 
43 
44 
43 
46 

47 
43 

49 

30 

5i 

52 

53 

54 

55 

56 

5 

5 

£ 

00 



8.766',- 5 
763323 
770070 

773101 
775223 

77-333 
779 '»34 
78i524 

- 36o5 

' J675 



8- 



8 



78-736 
789787 
791828 

798839 
795881 

79" 

799^97 

801892 

803876 

8o5 

807819 

771 

811-26 

8i36on 

81 5599 
817522 
819*36 
82i343 
828240 
825i3o 

■827011 
828884 
830749 
832607 
834456 
836297 
838 1. 3o 
880956 
841774 
843585 



4006 
3 9 -84 
3 9 -62 
39-41 

3 9 -ig 
38-98 

38-77 
38-57 

38-36 
33-i6 

37-96 

37.76 

3 7 -56 

3 7 -37 
37-n 
36- 9 8 

36 --9 
36-6i 
36-42 
36-24 
36- 06 

35-38 
35-70 
35-53 
35-35 
35.18 
35-oi 
34-34 
34-67 
34-31 
34-3i 

34-18 

34-02 
33-86 
33-70 
33-54 
33-39 
33-23 
33-o3 
32-93 
32-78 



Cosine 



32-63 

32-49 
32-34 

32-IQ 
32-03 

3i -91 
3i-77 
3i -63 
31-49 
3i-3a 

3l-22 

3i-08 

3o-q5 
3o-82 
30-69 
3o-56 
3o-43 
3o-3o 
3o-n 
3o-oo 



9-999404 
999398 
999891 
999384 

999378 

9998- r 

999364 

999 3 3" 
99g35o 

99934} 

999336 

9-999829 
909322 
9993 1 5 
999808 
999301 
999294 
999286 
999279 
999272 
999265 

9.999257 
999230 
999242 
999235 
999227 
007220 
999212 
999203 
999197 
999189 

9-999181 
999 1 74 
999 1 66 
999i53 

999IJO 

999 1 42 
999 1 34 
999126 
Q99I 1 3 
999 1 10 

9-999102 
999004 

999086 
999077 
999069 
99906 1 
999053 

999044 

,o36 

999027 

9.999019 
999010 
999002 
99^993 
99 

998976 
998967 
99^938 
99S930 
998941 



D. 



Sine 



8-719396 J 
72iSo6kJ 
724234 
726583 i 

728959 : 

73i3i7 

733663 1 
-35996 J 

738317 ! 
740626 ! 
742922 j 

8-745207 

7474-9 
749-40 
7 5i 9 Q 9 
754227 
756453 
758668 
76c - : 
-63o65 
763246 

3-76-417 
7693-3 
771727 
773S66 
775 99 5 

7781 14 
780222 
782820 
784408 
786486 

8-788554 
790613 
792662 
794701 
796781 
798752 
800763 
802765 
8o4758 
806742 

8-808717 
8io683 

8 1 264 1 
814389 
816529 
8 1 8461 
820334 
822298 
824205 
826103 

8-82-^092 
829 - 
83ii 
8336 1 3 

835471 
837321 
83gi63 
840998 
842823 
844644 



40-17 
39-93 
39-74 
39-52 
39-3o 
37-09 
33-89 
38-68 
38-48 
33-27 
38-o 7 

37-37 
37-68 

37-49 
37-29 
37-10 
36-92 
36-73 
36-55 
36-36 
36-iS 

36-00 
35-83 
35-65 
35-43 
35-3i 
35-i4 

34-97 
34- 80 

34-64 

34-47 

34-3i 
34-i5 
33-09 
33-83 
33-68 
33-52 
33- 3 7 

33-22 

33-07 
32-92 

32-78 

32-62 

32-43 
32-33 

32-10 

32-o5 
3 1 • 91 

3i-77 
3i-63 
3i-5o 

3 1 -36 

3i-23 
3i-io 
3o-o6 
3o-83 
3o--o 
3o-57 
3o-.',) 
3o-32 
3o-i9 



11 



\i 



L otang. 

1 1 • 280604 
2-8194 
2-3-96 
2-3412 
2-1041 
268( ^3 
266337 
264004 
2616^3 
2593-4 
257078 

254-98 

252521 
250200 
248ori 

243773 
243547 
241332 
239128 

236g35 
234734 

•232583 

230422 

2282-3 
226134 
224oo5 

221886 
210778 
217680 
215592 
2i35i4 

11-211446 

2oq - 
20-333 

2052Q9 
203269 
20I24O 
109231 
19-283 
1932 42 
193258 

)I233 
189817 
187359 

1 854i 1 
1 83471 

i8i53o 
1- .616 
177702 

I738Q7 

- 1-2008 
170126 

16' - 
164529 

i6r 
i6o33 7 
159002 
1 57175 
1 ).)356 



II-K 



II 



60 

5 9 

58 

57 
56 

55 

54 
53 

52 

5i 

30 

49 

43 

47 
46 
45 

I 43 

' 42 

41 

i 40 

|3 9 
33 

3 7 

I 36 

35 

34 
33 
32 
3i 



29 
28 
2- 
26 

25 
24 
23 
22 
21 
20 

10 

I- 

n 

16 
1 5 

14 
1 3 
12 
1 1 
10 

9 
8 

6 
5 

4 
3 
2 
I 
o 



Cotan?. 



I I). 



Taxis;. I M. 



(86 DEGREES.) 



22 



(4 DEGREES.) A TABLE OF LOGARITHMIC 



M. 




Sine 


D. 


Cosine 


D. 


Tang. 


D. 


Cotang. 




8-843585 


3o-o5 


9-908941 


• i5 


8 • 844644 


3o-i9 


n-155356 


60 


i 


' 845387 


29 


•92 


^998932 


• i5 


846455 


80-07 


153545 


5o 


2 


847i83 


29 


■80 


998923 


•i5 


848260 


29-95 


151740 


1 58 


3 


848971 


29 


•67 


998914 


-i5 


85oo57 


29-82 


140943 


5 7 


4 


85o75i 


29 


■ 55 


998905 


•i5 


85 1 846 


29-70 


1 48 1 54 


56 


5 


852525 


29 


•43 


998896 


-i5 


853628 


2 9 -58 


146372 


1 55 


6 


8:14291 


29 


•3i 


998887 


-i5 


8554o3 


29-46 


144597 


! 54 


7 


800049 


2Q 


• 19 


998878 


-i5 


85 7 i 7 i 


29-35 


142829 


53 


8 


837801 


20 


•07 


998869 


•i5 


858 9 32 


29-23 


141068 


52 


9 


859546 


28 


.06 


998860 


-i5 


860686 


29-11 


139314 


5i 


10 


86i283 


28 


• 84 


9 9 885i 


• i5 


862433 


29-00 


' 137567 


DC 


ii 


8-863oi 4 


28 


•73 


9-998841 


• i5 


8-864173 


28-88 


ii-i35827 


4q 


12 


864738 


28 


•61 


99 8832 


-i5 


865 9 o6 


28-77 


134094 


48 


i3 


866455 


28 


•5o 


998823 


.16 


867632 


28-66 


132368 


47 


14 


868i65 


28 


-3o 


9 9 88i3 


.16 


869351 


28.54 


130649 


46 


i5 


869868 


28 


■ 28 


998804 


.16 


871064 


28.43 


I28 9 36 


45 


16 


87i565 


28 


■17 


998795 


.16 


872770 


28.32 


127230 


44 


n 


873255 


28 


• 06 


998785 


.16 


874469 


28-21 


I2553i 


43 


18 


874938 


27 


- 9 5 


998776 


.16 


876162 


28.11 


123838 


42 


19 


876615 


27 


• 86 


998766 


• 16 


877849 


28-00 


I22l5l 


4i 


20 


878285 


27 


- 7 3 


998757 


• 16 


879529 


27-89 


1 20471 


4o 


21 


8-879949 


27 


■ 63 


9-998747 


•16 


8-881202 


27.79 


11-118798 


3 2 


22 


881607 


27 


•52 


998738 


•16 


882869 


27-68 


H7i3i 


38 


23 


883258 


27 


• 42 


998728 


.16 


88453o 


27-58 


1 1 5470 


37 


24 


884oo3 


27 


■3i 


998718 


•16 


886i85 


27-47 


n38i5 


36 


25 


886042 


27 


•21 


998708 


•16 


887833 


27-37 


112167 


35 


26 


888174 


27 


•11 


998699 


•16 


889476 


27-27 


no524 


34 


27 


889801 


27 


• 00 


998689 


•16 


891 112 


27-17 


108888 


33 


28 


89 1 42 1 


26 


• 90 


998679 


•16 


892742 


27-07 


107258 


32 


29 


8 9 3o35 


26 


• 80 


998669 


•17 


8 9 4366 


26-07 


io5634 


3i 


3o 


894643 


26 


.70 


998659 


•17 


8 9 5 9 84 


26-87 


104016 


3o 


3i 


8-896246 


26 


• 6o 


9 • 998649 


•n 


8-897596 


26-77 


11-102404 


29 


32 


807842 


26 


• 5i 


998639 


•17 


899203 


26-67 


100797 


28 


33 


899432 


26 


.41 


998629 


•17 


900803 


26-58 


099197 


27 


34 


90 1 017 


26 


3i 


998619 


•*7 


902398 


26-48 


097602 


26 


35 


902596 


26 


•22 


998609 


•17 


903987 


26-38 


096013 


25 


36 


904169 


26 


•12 


998599 


•17 


905570 


26-29 


094430 


24 ' 


37 


905736 


26 


o3 


998589 


•17 


907147 


26-20 


092853 


23 


38 


907297 


25 


2 3 


998678 


•17 


908719 


26-10 


091281 
089715 


22 


3 9 


908853 


25 


84 


99 8568 


•17 


910280 


26-01 


21 


4o 


9 1 0404 


25 


75 


99 8558 


•17 


91 1 846 


25-92 


o88i54 


20 


4i 


8-911949 


25 


66 


9-998648 


•H 


8-913401 


25-83 


11-086599 


IO 


42 


913488 


25 


56 


998537 


•n 


9 1 495 1 


25-74 


085049 


10 


43 


915022 


25 


47 


998627 


•17 


916495 


25-65 


o835od 


17 


44 


9i655o 


25 


38 


998516 


•18 


qi8o34 


25-56 


081966 


IO 


45 


918073 


25 


29 


998606 


•18 


919568 


25-47 


080432 


i5 


46 


919691 


25 


20 


998495 
998485 


•18 


921096 


25-38 


078904 


14 


47 


921103 


25 


12 


•18 


922619 


25-3o 


077381 


i3 


48 


922610 


25 


o3 


998474 


•18 


924i36 


25-21 


075864 


12 


49 


924112 


24 


94 


998464 


•18 


925649 


25- 12 


07435i 


1 1 


5o 


925609 


24 


86 


998453 


•18 


927156 


25 -o3 


072844 


10 


5i 


8*927100 ' 


24- 


77 


9-998442 


.18 


8-928658 


24 -oi 


11-071342 





52 


928587 


24 


69 


998431 


• 18 


93oi55 


24-86 


069845 





53 


930068 


24- 


60 


998421 


.18 


931647 


24-78 


o68353 


7 


54 


93 1 544 


24" 


52 


998410 


.18 


9 33i34 


24-70 


066866 


6 


55 


933oi5 


24- 


43 


998399 


.18 


934616 


24-6i 


065384 


5 


56 


934481 


24- 


35 


9 9 8388 


.18 


936093 


24-53 


063902 


4 


57 


935942 


24- 


27 


998377 


.18 


937565 


24-45 


06243 D 


3 


58 


937398 


24- 


*9 


99 8366 


• 18 


939032 


24-3 7 


060968 


2 


5 9 


93885o 


24- 


11 


998355 


•18 


940494 
941952 


24-3o 


o5o5o6 


1 


60 


940296 


24 -o3 j 


998344 


•18 


24-21 


o5oo48 




M. 
..... 


1, 


Cosine 


D. I 


Sine 




Cotang. 


D. 


Tang. 



(85 DEGREE^ 



SINES AND TAXGE^TS. (5 DEGKEES.) 



2d 



u. 



Sino 



o 
i 

2 

3 

4 

5 
6 

7 
8 

9 
lo 

H 

12 

13 

M 

i5 
16 
l 
i 

l 9 

20 

21 
22 
23 

24 
25 
26 

u 
11 

3i 

32 

33 

34 
35 
36 
3 

3 

3 9 
4o 

4i 

42 

43 
44 
45 
46 

4 

5o 
5i 

52 

53 
54 
55 
56 
5 

h 



8-94c?^6 
94 /38 
943174 
944606 

946034 
9474-6 
94887 i 

9)0287 
93 1 696 
q5; 1 00 

9 "'4499 

8 955H94 

9572-4 
95b6^o 
9600:12 
961429 
962801 
964170 
965534 
966893 
968249 

y. 069600 

97°';47 

9722^9 

973628 
974962 
976293 
•977619 
978941 
98025 
9 8i5 7 

8-982883 
984189 
985491 

9S6789 



D. 



989374 
990660 
991943 
993222 

994497 
8-995768 
9970 36 
998299 
999660 
9-000^16 

OO2069 

oo3.i 1 8 
0O4563 

or , J8o3 
007044 

9-008278 
0095 1 
010737 
01 1962 
01 3 182 
014400 
01 56i 3 
016824 
oi8o3i 
019235 



24 -o3 
23- 9 4 
23-87 
23=79 
23-71 
23-63 
23-55 
23-48 
23-4o 

23-32 
23-25 

23-17 

23-10 
23-02 

22-93 

22-88 
22-80 

22-73 

22-66 
22-59 
22-52 

22.44 

22-38 
22-3l 
22-24 
22-17 
22- 10 
22 -03 

21-97 

21 

21 



Cosine 



1 -30 

n-83 



21 

21 
21 
21 
21 
21 
21 



77 
70 

63 

5 7 

5o 

44 

38 



21 -3i 
21 -25 



21 



19 
12 
06 
00 



21 
21 
21 

20- 
2C 

20-76 
20-70 
20-64 

58 



20-94 

20-87 



20 

20 
20 
20 
20 
20 
20 
20 
20 
20 
20 



52 

46 
40 

34 

29 

23 

17 

12 
06 
00 



D. 



Tun£. 



9-998344 
998333 
998322 
9983 1 1 
998300 
998289 
998277 
998266 
998255 
998243 
998232 

9.99S220 
998209 
998107 
998186 
998174 
998163 
998i5i 
998139 
998128 
9981 16 

9-998104 
998092 
998080 
998068 
998056 

998044 
998032 
998020 
998008 
997996 

9.997985 
997972 
997959 

997947 
997933 
997922 
997910 
997897 
997885 
997872 

9-997860 

997847 
997835 
997822 
997809 

997797 
997784 

997771 
997758 
997743 

)- 997732 

997719 
997706 

997693 
997680 
997667 
997654 
997641 
997628 
997614 



!9 

19 

*9 
*9 

19 
19 
] 9 
'9 
*9 
J9 
J 9 

19 
l 9 

*9 

19 

19 

19 
20 

20 

20 

20 
20 
20 
20 
20 
20 
20 
20 
20 
20 

20 
20 
20 
20 

21 
21 
21 
21 
21 
21 

21 
21 
21 
21 
21 
21 
21 
21 
21 
21 

21 
21 
21 
22 
22 
22 
22 
22 
22 
22 



D. 



Cotang. 



8-941952 
943404 
944852 
946295 

947734 
949168 
930397 
952021 

933441 
934836 
936267 

8-937674 
959075 
960473 
961866 
963255 
964639 
966019 
967394 
968766 
970i33 

8-971496 
972835 
974209 
97556o 
976906 
978248 
979586 
980921 
982251 
983577 

3-984899 
986217 
987532 
988842 
990149 
99i45i 
992750 
994o45 
995337 
996624 

997908 
999188 
9-ooo465 
001738 
003007 
004272 

O03334 

006792 
008047 
009298 

9-010546 
01 1790 
oi3o3i 
014268 
oi55o2 
016732 
01 79 5o 
01918} 
020403 
021620 



8 



Cosine 



D. 



Sine 



Cotang . 



24-21 
24-i3 
24-o5 
23-97 
23-9C 

23-82 

23-74 

23-66 
23 -6c 

23-01 

23-44 

23-3 7 
23 -2g 
23-23 

23-14 

23-07 

23-00 
22-o3 
22-86 

22-79 
22-71 

22-65 
22-57 
22-5l 
22-44 
22-37 
22-3(2 
22-23 
22-17 
22- 10 
22-04 

21-97 

21 -91 

21-84 
21-78 
21 -71 
21 -65 
21-58 
21 
21 
21 



1 1 



52 

.46 
.40 



■ 27 

■21 

■i5 
.09 
■o3 



21 
21 
21 
21 
21 
21 
20-97 

20-OI 

20-85 

20- 80 

20-74 
20-68 
20-62 
20-56 

20-5l 

20-45 
20-40 
20-33 
20-28 

20-23 



II 



ill 



D. 



■o58o43 
056396 
o55i4S 
o537o5 
o52266 
o5o832 
049403 

o47979 
046539 

o45i44 
043733 

•042326 
040915 
039^27 
o3Si34 
036743 
o3536i 
o33q8i 
o326o6 
o3i234 
029S67 

■O285o4 
027145 
025791 
024440 
023094 
021752 
020414 
019079 

017749 
016423 

it -oi5ioi 
013783 
012468 
01 1 1 58 
0098 5 1 
ooo549 
007250 
oo5955 
oo4663 
003376 

11 -002092 
000812 
999535 
998262 
996993 
993728 
994466 
993208 
991933 
990702 

10-989454 
9SS210 
986969 
985732 

9 s W98 

983268 

982041 
980^ 1 7 
979597 
978380 



60 



10 



37 
56 
55 

54 
53 

52 
31 
DO 

4Q 
AS 

47 
46 
45 
44 
43 
42 
41 
40 

39 

38 

^7 
36 

35 

34 
33 

32 

3i 

3o 

29 
28 
27 
26 

25 

24 

23 

22 
21 

20 

18 

17 

16 

i5 

14 
i3 
12 
1 1 
10 



Tang. M 



7 
6 
5 
4 
3 
2 
1 
o 



(84 DEGREES.) 



24 


(6 : 


DEGREES.) A TABLE OF LOGARITHMIC 




M. 


Sine 


D. 


Cosine 


D. 


Tang. 


D. 


Colang. 




o 


9-019235 


• 20-00 


9.9976U 


•22 


9-021620 


20-23 


10-978380 J 


60 


i 


020435 


I9-95 


997601 


•22 


022834 


20>I7 


977166 1 3Q 


2 


021632 


19-89 


997588 


•22 


024044 


20-11 


9769,56 


58 


3 


022825 


I9.84 


997574 


•22 


02525l 


20- 06 


9747t9 


57 


4 


024016 


I9.78 


997561 


•22 


026455 


20-00 


973543 


56 


5 


025203 


I9-73 


997547 


•22 


027655 


I9.95 


972345 


55 


6 


026386 


19-67 


997534 


•23 


028852 


I9.9O 


971 148 


54 


7 


027567 


19-62 


997520 


•23 


o3oo46 


19-85 ; 


969954 
968763 


53 


8 


028744 


I9-57 


997507 


•23 


o3i237 


19.79 


52 


9 


029918 


I9-5i 


997493 


•23 


o32425 


19-74 


967575 


5i 


10 


031089 


19-47 


997480 


•23 


033609 


19-69 


96639I 


5o 1 


ii 


9«o32257 


19-41 


9-997466 


•23 


9-034791 


19-64 


10-965209 


49 


12 


o3342i 


19-36 


997452 


•23 


035969 


i 9 -58 


96403 1 


48 


i3 


o34582 


i9-3o 


997439 


•23 


037144 


19-53 


962856 


47 


14 


035741 


19-25 


997423 


•23 


o383i6 


19.48 


961684 , 


46 


15 


o368 9 6 


19-20 


99741 1 


•23 


039485 


19.43 


9605 1 5 


45 


16 


o38o48 


19. i5 


997397 


•23 


0406 5 1 


*9-?? 


939349 


44 


l l 


039197 


19. 10 


997383 


•23 


041813 


19-33 


958187 


43 


18 


040342 


lQ-05 


907369 


•23 


042973 


19-28 


957027 


42 


'9 


o4i485 


18-99 


997355 


•23 


0441 3o 


19-23 


935870 


41 


20 


042625 


18.94 


997341 


•23 


045284 


19-18 


954716 


40 


21 


9-043762 


18-89 


9.997327 


•24 


9-046434 


19. i3 


10.953566 


3o 


22 


• 044893 


18-84 


9973 1 3 


•24 


047582 


19.08 


952418 


38 


23 


046026 


18-79 


697299 


•24 


048727 


19 -o3 


931273 


37 


24 


o47i54 


18.75 


997283 


•24 


049869 


16.98 


95oi3i 


36 


25 


048279 


18-70 


997271 


•24 


o5iooo 


i8- 9 3 


948992 


35 


26 


049400 


18. 65 


997257 


•24 


o52i44 


18-89 


947856 


34 


27 


o5o5i9 


18-60 


997242 


•24 


053277 


18-84 


946723 


33 


28 


o5i635 


18-55 


997228 


•24 


054407 


18.79 


945593 


32 


29 


052749 


18. 5o 


997214 


.24 


o55535 


18.74 


944465 


3i 


3o 


o5385 9 


18-45 


997199 


•24 


o56659 


18-70 


943341 


3o 


3i 


9-054966 


18.41 


9.997185 


•24 


9-057781 


i8-65 


10-942219 


29 


32 


056071 


18. 36 


997170 


•24 


058900 


18-69 


941 100 


28 , 


33 


057172 


18. 3i 


997 1 56 


•24 


060016 


18.55 


930984 


27 


34 


058271 


18-27 


997141 


•24 


061 i3o 


18. 5i 


938870 


26 


35 


059367 


18-22 


997127 


•24 


062240 


18-46 


937760 


25 


36 


060460 


18-17 


9971 12 


.24 


o63348 


18-42 


936652 


24 


37 


o6i55i 


i8-i3 


997098 


•24 


o64453 


18-37 


935547 


23 


38 


062639 


18-08 


997083 


•25 


o65556 


i8-33 


934444 


22 


3 9 


063724 


18-04 


997068 


•25 


o66655 


18-28 


933345 


21 


4o 


064806 


17-99 


997053 


•25 


067752 


18-24 


932248 


20 


4i 


9-o65885 


I7;94 


9-Q97039 


•25 


9-068846 


18-19 


io-93ii54 


IO 


42 


066962 


17*90 


997024 


•25 


069938 


i8-i5 


930062 


10 


43 


o68o36 


17-06 


997009 


•25 


071027 


i8-ic 


928973 


17 


44 


069107 


17.81 


996994 


•25 


072113 


18-06 


927887 


16 


45 


070176 


n-ii 


996979 


•25 


073197 


iS-QS 


926803 


i5 


46 


071242 


17.72 


996964 


•25 


074278 


17.97 


925722 


14 


% 


072306 


17-68 


996949 


•25 


075356 


I7.93 


924644 


i3 


073366 


i 7 -63 


996934 


•25 


076432 


17.89 


9 23568 


12 


49 


074424 


i 7 -5o 


996919 


•25 


077505 


17-84 


922495 


1 1 


5o 


075480 


17-55 


996904 


•25 


078576 


17-80 


921424 


10 


5i 


0-076533 


17 «5o 


9-996889 


•25 


9 • 079644 


17.76 


10-920356 


Q 


5a 


077583 


17-46 


996874 


• 25 


080710 


17-72 


919290 


8 


53 


078631 


17-42 


9 9 6858 


•25 


081773 


17.67 


918227 


7 


54 


079676 


17-38 


996843 


•25 


o82833 


17-63 


917167 


b 


55 


080719 


i 7 -33 


996828 


•25 


o838 9 i 


17.59 


9 161 09 


5 


56 


081759 


17-29 


996812 


.26 


084947 


17 53 


9i5o53 


4 


58 


082797 


17-23 


996797 


• 26 


086000 


17. 5i 


914000 


3 


083832 


17-21 


996782 


.26 


087050 


17-47 


912950 


2 


5 9 


084864 


17-17 


996766 


.26 


088098 


J7-43 


91 1902 


1 


66 


085894 


17- 13 


996751 


• 26 


089144 


17-38 


9io856 



M. 


i 


Cosine 


1). 


Sine 


Cotansr. 

P , - 


D. 


_... Tan ff- 



(83 DEGREES.) 



SINES AND TANGENTS, (7 DEGEEES.) 



25 



ML. 



Sine 



D. 



Cosine 






9-085894 


I 


086922 


2 


087947 


3 


088970 


4 


089990 


5 


091008 


6 


092024 


I 


093037 


094047 


9 


o9§o56 


10 


096062 


ii 


9-097065 


12 


098066 


i3 


099065 


14 


100062 


15 


ioio56 


16 


102048 


17 


io3o37 


18 


io4o25 


»9 


ioSoio 


20 


105992 


21 


9-106973 


22 


107951 


23 


108927 


24 


1 0990 1 


20 


110873 


26 


11 1842 


27 


1 1 2809 


28 


113774 


29 


II4737 


3o 


1 1 56 9 8 


3i 


9- 1 1 6656 


32 


1 1 76 1 3 


33 


1 18D67 


34 


119519 


35 


120469 


36 


121417 


37 


122362 


38 


l233o6 


3q 


124248 


40 


125187 


41 


9-126125 


42 


127060 


43 


127993 


44 


128025 
129854 


45 


46 


130781 


a 


131706 


i3263o 


49 


i3355i 


5o 


134470 


51 


9-135387 


52 


i363o3 


53 


137216 


54 


138128 


55 


139037 


56 


139944 
i4o85o 


n 


i4n 5 4 


5o 


142655 


60 


143555 



17- 13 

17.09 
17.04 
17-00 
16 96 
16-92 
16.88 
16-84 
16-80 
16-76 
16-73 

16-68 
i6-65 
16-61 
16-57 
i6-53 
16-49 
i6-43 
16-41 
i6-38 
16-34 

i6-3o 
16-27 
16-23 
16-19 
l6-l6 
16-12 
16-08 
16 -o5 
16-0: 
i5-97 

i5-94 
1 5- 00 
i5-87 
15-83 
i5-8o 
15-76 
15-73 
1 5 -69 
i5-66 
i5>62 

15.59 
i5-56 
i5-52 
15.49 
i5-43 
15-42 
15.39 
i5-3o 
i5-32 
15-29 

i5-25 

l5-22 

i5- 19 

i5-i6 

l5-12 

15.09 

l5-06 

i5-o3 
i5oo 

14-9° 



D. 



Tans 



D. 



Cotang. 



9.996751 
996735 
996720 
996704 
996688 
996673 
996607 
996641 
996625 
996610 
996094 

9.996578 
996562 
996046 
996030 
996514 
996498 
996482 
996465 

996449 
996433 

9.996417 
996400 
996384 
99 6368 
99&35i 
996335 
996318 
996302 
996286 
996269 

9-996252 
996235 
996219 
996202 
996185 
996168 
996 1 5 1 
996134 
996 1 1 7 
996100 

9.996083 
996066 
996049 
996032 
9960 1 5 
990998 
995980 
99O963 
9909I6 
995928 



9-9959 

)58 



11 

99 58 9 4 
995876 
995859 
995841 
990823 
995806 
995788 

995771 

995753 



.26 
.26 

.26 
.26 

• 26 

• 26 
•26 
.26 
.26 
•26 
.26 

•27 

•27 

•27 

•27 
.27 

•27 
.27 

•27 
•27 
•27 

•27 
•27 
•27 
•27 
•27 

•27 
•27 
.28 
.28 
•28 

■ 28 
.28 
.28 
.28 
.28 
.28 
.28 
.28 
.28 
.28 

•29 

•29 

•29 

•29 
.29 

•29 
.29 

•29 

•29 

•29 

.29 

.29 

.29 

■29 

•29 

■29 
.29 

•-9 

•29 
.29 



9-089144 
090187 
091228 
092266 
093302 
094336 
095367 
096395 
097422 
098446 
099468 

9.100487 
ioi5o4 
102519 
io3532 
104542 
io555o 
io6556 
107559 
io856o 
109559 

'iio556 
iii55i 
1 12543 

1 1 3533 
ii452i 
n55o7 
116491 
117472 
1 i8452 
119429 

•120404 
1 21377 
122348 
I233i7 
124284 
125249 
126211 
127172 
i28i3o 
129087 

•i3oo4i 
1 30994 
131944 

i328 9 3 
i3383g 
134784 
135726 
136667 
137605 
138542 

1 -139476 
140409 
i4i3\4o 
142269 
143196 
144121 
i45o44 
145966 
146885 
147803 



17-38 
17-34 
17.30 
17-27 
17-22 
17-19 
17-10 
17-11 
17-07 
17 -o3 
16.99 

16-95 
16.91 
16-87 
16-84 
16-80 
16-76 
16-72 
16-69 
16-60 
16-61 

16. 58 
i6-54 
i6-5o 
16-46 
i6«43 
16.39 
i6- 36 
16-32 
16-29 
16-20 

16-22 

16.18 

i6- 15 
16- 11 
16-07 
16-04 
16-01 
15.97 
i5-94 
15.91 

15-87 
15-84 
i5.8i 
15-77 
i5-74 
15.71 
i5-6 7 
i5-64 
i5-6i 
15-58 

i5-55 

i5-5i 

i5-48 

i5-45 

i5-42 

i5-39 

15.35 

i5-32 

i5-29 

i5-26 



"1 



10-910856 

909813 
908772 
907734 
906698 
905664 
904633 
9o36o5 
902578 
90i554 
90c532 

ji 0-8995 1 3 
898496 
897481 
896468 
8 9 5458 
894450 
893444 
892441 
891440 

890441 

10-889444 
888449 
887457 
886467 
885479 
884493 
883509 
882526 
881548 
880571 

10-879596 
878623 
877652 
876683 
875716 
8i475i 
873789 
872828 
871870 
870913 

10-869959 
869006 
868o56 
867107 
866161 
8652i6 
864274 
863333 
862395 
86i438 

io-86o524 
85 9 59i 
85866o 
85773i 

8568o4 
855879 
854956 
854o3/» 
853n5 
852197 



Cosine 



D. 



Bine 



Cotang. 



D. 



60 

57 
56 
55 

54 
53 
5a 
5i 
5o 

49 

48 

47 
46 
45 
44 
43 
42 
4i 
4o 

3g 
38 

37 
36 

35 

34 
33 

32 

3i 
3o 

29 
28 

27 
56 

25 

24 

23 
22 
21 

20 

19 

3 

i5 
14 
i3 
12 
11 
10 



Toy&_ 



5 

4 
3 
2 

1 
o 



M. 



(82 DEGREES.) 



26 



(8 DEGEEES ) A TABLE OF LOGARITHMIC 



M. 



Sine 



D. 



o 
i 

2 

3 

4 
5 
6 

7 
8 

9 

10 

n 

12 

i3 

U 
i5 
16 

H 
18 

19 

20 

21 
22 
23 

24 
25 

26 

27 
28 

3o 
3i 

32 

33 

34 
35 
36 

37 
38 

39 

40 

41 
42 
43 
44 
45 
46 

4 

5o 
5i 

52 

53 

54 

55 

56 

5 

5 

59 

60 



9. 



9- 



43555 

44453 
4534Q 
46243 
47i36 
48026 
480 1 5 
49802 
5o686 
5 1 369 
5245i 

5333o 
54208 
55o83 
55o5 7 
5683o 
57700 
5856 9 
5 9 435 
6o3oi 
61164 

62025 
62885 
63743 
64600 
65454 
66307 
67159 
68008 
68856 
69702 

70547 
7i38 9 
72230 
73070 
73908 

74744 
75578 
7641 1 
77242 
78072 

78900 
79726 
8o55i 
81374 
82196 
83oi6 
83834 
8465 1 
85466 
86280 

87092 
87903 
88712 
89519 

90323 

9ii3o 
91933 
92734 
9 3534 
94332 



Cosine 



14-96 
14-93 
14-qo 
14-87 
14-84 
14-81 
14-78 
14-75 
14-72 
14-69 
14-66 

14-63 
14-60 
14-57 
14-54 
i4-5i 
14-48 
U-45 
14-42 
14-39 
14-36 

14-33 
l4»3o 
14-27 
14-24 

14-22 
14-19 

14-16 
i4-i3 
14-10 

14-07 

l4-o5 
14-02 
i3-99 
13-96 
13-94 
13-91 
13-88 
13-86 
13-83 
r3-8o 

13-77 

13-74 
i3-72 
13-69 
i3-66 
i3-64 
i3-6i 
i3-5g 
13-56 
13-53 

135.1 

13-48 
i3-46 
i3-43 
i3-4i 
i3-38 
13-36 
i3-33 
i3-3o 
i3-28 



Cosine 



D. 



9 995753 

995735 
995717 

99 5o 99 
995681 

995664 
995646 
995628 
995610 
995591 
995573 

9-995555 
995537 
995519 
9955oi 
995482 
995464 
995446 
995427 
995409 
995390 

9-995372 
995353 
995334 
9953i6 
995297 
995278 
995260 
995241 
995222 
9952o3 

9-995184 
995i65 
995146 
995127 
995108 
995089 
995070 
995o5i 
995o32 
995oi3 

9.994993 

994974 
994955 
994935 
994916 
994896 
994877 
994857 
994838 
994818 

9.99479 s 
994779 
994759 
994739 

994719 
994700 

994680 

994660 

994640 

994620 



D. 



Sine 



3o 
3o 
3o 
3o 
3o 
3o 
3o 
3o 
3o 
3o 
3o 

3o 
3o 
3o 
3i 
3i 
3i 
3i 
3i 
3i 
3i 

3i 

3i 
3i 
3i 
3i 
3i 
3i 

32 
32 

32 

32 
32 
32 
32 
32 
32 
32 
32 
32 
32 

32 
32 
32 
32 

33 
33 
33 
33 
33 
33 

33 
33 
33 
33 
33 
33 
33 
33 
33 
33 



Tang. 



9- 



9- 



47803 
48718 
49632 
5o544 
5i454 
52363 
53269 

54174 
5507- 
55978 
56877 

57775 
586 7 i 
59565 
60457 
6i347 
62236 
63i23 
64008 
64892 
65774 

66654 
67532 
68409 
69284 
70157 
71029 
71899 
72767 
73634 
74499 
7 5362 
76224 
77o 8 4 
77942 
78799 
79655 
8o5o8 
8i36o 
82211 
83o59 

83907 
84752 
85597 
864I9 
87280 
88120 
88 9 58 

89794 
90629 
91462 

92294 
93124 
93953 
9 1780 
95606 
96430 
97253 
98074 
98894 
99713 



D. 



i5-26 
i5-23 

l5-20 

i5-i7 
i5-i4 
iS-ii 

i5-o8 
i5-o5 

l5-02 

14-99 
14-96 

I4-93 
14-qo 

14-87 
14-84 
14-81 
14-79 
14-76 
14-73 
14-70 
14-67 

14-64 

14-61 
14-58 
14-55 
14-53 
i4-5o 

14-47 
14-44 
14-42 
14-39 

14-36 
U-33 
i4-3i 
14-28 
14-25 
14-23 
14-20 

14-17 
i4-i5 
14-12 

14-09 
14-07 
14-04 
14-02 
i3-99 
13.96 
13.93 
i3-oi 
13.89 
13-86 

i3-84 
i3-8i 
13.79 
13.76 
i3-74 
i3 • "7 1 
i3-09 
13-66 
13-64 
i3-6i 



Cotang. 



10 



•852197 
851282 
85o368 
849456 
848546 
847637 
846731 
. 845826 
844923 
• 844022 
843i23 

10-842225 
841329 
840435 
839543 
838653 
837764 
S36877 
835 99 2 
835io8 
834226 

10-833346 
832468 
83i5 9 i 
830716 
829843 
828971 
828101 
827233 
826366 
8255oi 



10 



10 



Cotang. 



D. 



824638 
823776 
822916 
822058 
821201 
820345 
819492 
818640 
817789 
816941 

816093 
8i5248 
8i44o3 
8i356i 
812720 
811880 
81 1042 
810206 
809371 
8o8538 

10-807706 

806876 
806047 

8o5220 

804394 
803570 

803747 
801926 
801 106 

800287 



Tang. 



60 

58 

57 
56 
55 
54 
53 

52 

5i 

5o 

49 
48 

47 
46 
45 
44 
43 
42 
4i 
4o 

39 
38 

37 
36 
35 

34 

33 

32 

3i 
3o 

29 

28 

27 
26 

25 
24 
23 
22 
21 
20 

19 
13 
17 
16 

i5 
i4 
i3 
12 
11 
10 



7 
6 
5 
4 
3 
2 
1 
o 



(81 DEGREES.) 



SINES AND TANGENTS. (9 DEGREE.) 



27 



M. 


Sine \ 


D. 


o 


9-194332 


13-28 


1 

i 


195129 J 


i3-26 


2 


193920 j 


i3-23 


3 


196719 1 


l3-2I 


4 


197011 


i3-i8 


5 


198302 


i3-i6 


6 


199091 


i3-i3 


7 


199S79 


1 3 - 1 1 


8 


200666 


i3-o8 


9 


201401 


i3-o6 


10 | 


202234 


i3-04 


n 


9-203017 


i3-oi 


12 


203797 


12-99 


i3 


204077 


12-96 


M 


205354 


12-94 


i5 


2o6i3i 


12-92 
12-89 


16 


206906 


i? 


207679 


12-87 


18 


208402 


12-85 


J 9 


209222 


12-82 


20 


209992 


12 -8o 


21 


9-210760 


12-78 


22 


2Il5?6 


12-75 


23 


2I2SQI 


12-73 


24 


2i3o55 


12-71 


25 


2i38i8 


12-68 


26 


214379 
' 2i5333 


12-66 


3 


12-64 


216097 


I2-6l 


2 9 


2i6854 


12-59 


3o 


2 1 7609 


12-57 


3i 


9-2i8363 


12-55 


32 


2191 16 


12-53 


33 


219868 


I2-5o 




220618 


12-48 


35 


221 367 


12-46 


36 


2221 15 


12.44 


37 


222861 


12-42 


38 


223606 


12-39 


3 9 


224349 


12-37 


4o 


225092 


12*35 


4i 


9-225833 


12-33 


42 


226573 


12 -3 1 


43 


22731 1 


12-28 


44 


228048 


12-26 


45 


228784 


12-24 


46 


229518 


12-22 


47 


2lo2")2 


12-20 


48 


230984 


12-18 


49 


23l7l4 


12 • 16 


5o 


2 32444 


I2-I4 


Si 


9-233172 


12-12 


52 


2338 9 9 


, I 2 • 09 


53 


2 3162 


I 12 -°] 


54 


235349 
236073 


12-05 


55 


12-o3 


56 


236795 


12-01 


57 


2 3 7 5 1 5 


11.99 


58 


a3«a35 


u-97 


5 9 


238 9 53 


11 -90 


6o 


23967c 


11-93 


I 

l 


Cosine 

— lb — 


1 I>- 



Cosine 

9-994620 
994600 
994580 
99456o 
994540 
994019 

994499 
9944-9 
994409 
99-^438 

994418 

9-994397 

9943-7 
994357 
994336 
9943i6 
994295 
904274 
9942 54 
994233 
994212 

9-994191 
9941 7 1 
9941 5o 
994129 
994108 
994087 
994066 
994045 
994024 
994oo3 

993981 
993960 
9 9 3 9 3o 
9939,8 
993-190 
993875 
993854 
9 9 3832 
99381 1 
993789 

.993768 
993746 
993725 
993703 
99 3 68 1 
993660 
993638 
993616 
993594 
993572 

•99355o 
993528 
9935o6 
993484 
993462 
993440 
993418 
993396 

993374 
99335i 



D. 



Tang. 



D. 



• 33 

• 33 

• 33 

34 
•34 
•34 
•34 
•34 
•34 
•34 
•34 

•34 
•34 
•34 
•34 
•34 
•34 
•35 
•35 
■35 

• 35 

• 35 

• 35 
•35 
•35 

• 35 

• 35 

• 35 
•35 

• 35 
•35 

• 35 
•35 

• 35 
•35 

• 36 

• 36 

• 36 

• 36 
•36 

• 36 

•36 
•36 
■36 
■36 

• 36 

• 36 

• 36. 
•36 
•37 
•37 

•37 
.37 

•37 

•37 
•3 7 
•37 
.37 
.37 
• 3 7 
•37 



9-199713 

200529 
201345 
202109 
202971 
203782 
204092 
205400 
206207 
207013 
207817 

9.208619 
209420 

210220 
2lI0l8 

2ii8i5 
212611 
2i34o5 
214198 
214989 
210780 

9-2i6568 
217356 

218142 
218926 
219710 
220492 
221272 
222002 
222830 
2236o6 

1-224382 

225i56 
225929 
226700 

227471 
228239 
229007 
229773 
23o539 
23i3o2 



9- 



232o65 

232^26 

2 33586 
234345 
235io3 
23585g 
2 366 1 4 
23 7 368 

238l20 

238872 

23962 i 

240371 
241118 

24i865 
242610 
243354 
244097 
2 i ',8 i 9 

245579 

246319 



i3-6i 
i3 -69 

i3-56 
i3-54 
13-52 
i3-49 
13-47 
i3-4o 
13-42 
i3-4o 
13-38 

i3 • 35 

i3-33 
i3-3i 
i3-28 

i3-26 
i3-24 

l3-2I 

i3 - 19 

I3-I7 
i3-i5 

I3-I2 

i3-io 
i3-o8 
i3-o5 
i3-o3 
i3-oi 
12-99 
12-97 
12-94 
12-92 

12-90 
12-88 
12-86 
12-84 
12-81 
12-79 



Cotang. 



12 
12 
12 
12 

12 
12 
12 
12 
12 



77 
7^ 

73 
7i 

69 
.67 
-65 
• 62 
.60 
12-58 
12-56 
1 2 • 54 

12-52 
12 -5o 

12-48 

12-46 

12-44 

12-42 
12-40 

12-38 
12-36 

12-3'* 
12-32 

i2-3o 



10-800287 

799471 

798655 

797841 
797029 
796218 
795408 
794600 
793793 

792987 

792183 

io-79i38i 
790580 
789780 
788982 
788180 
7873S9 
786090 
785802 
785oi 1 
784220 

10-783432 
782644 
78i858 
781074 
780290 
779508 
778728 
777943 
777170 
7763 9 4 

10-775618 
774844 
774071 
7733oo 
772029 
771761 
770993 
770227 
769461 
768698 

10-767935 



76717. 

766414 

765655 

764897 

7641 ii 
763386 
762632 
761880 
761:28 

.760378 
759629 
758882 
7681 35 
757390 
756646 
705903 
755 161 

754 ',2 1 
7 5368l 



I 

5 
4 
3 
2 
1 
o 



Sine 



Cotang. 



D. 



Tang. 1 &1 



(80 DEGREES.) 



28 


(10 


DEGREES.) A TABLE OF LOGAI 


IITD 


MIC 




M. 


Sine 


D. 


Cosine 


D. 


Tang. £>. 


Cotang. 


60 


o 


9-239670 


11-93 


9'99335i 


• 3 7 


9 -2463 19 12 • 


3o 


10-753681 


i 


24o386 


11.91 


993329 


•3? 


247057 12- 


28 


752943 


5o 


2 


241 101 


11-89 


993307 


.37 


247794 J2- 


26 


7*2206 


58 


3 


241814 


11-87 


993285 


• 3 7 


24853o 12 • 


24 


751470 


57 


4 


242526 


n-85 


9Q3262 


• 3 7 


249264 12- 


22 


750736 . 56 


5 


243237 


n-83 


993240 


• 3 7 


249998 12- 


20 


750002 1 55 


6 


243947 


11.81 


993217 


• 38 


25o73o 12- 


18 


7492*70 


04 


H 


244656 


11.79 


993195 


• 38 


25i46i 12- 


H 


748539 


53 


8 


245363 


n-77 


993172 


• 38 


252I9I 12- 


i5 


747809 


52 


9 


246069 


ii- 7 5 


993 1 49 


• 38 


252920 12- 


i3 


747080 


5i 


10 


246776 


11-73 


993127 


• 38 


253648 12- 


11 


746352 


5o 


ii 


9' 247478 


11-71 


9 • 993 i 04 


• 38 


9-254374 12 


09 


10-745626 


40 


12 


248 1 81 


11-69 


993081 


■ 38 


255lOO 12 


3 


744900 


48 


i3 


248883 


11-67 


993059 


• 38 


255824 12 


744176 


47 


u 


249583 


n-65 


993o36 


• 38 


256547 12 


o3 


743453 


46 


i5 


250282 


u-63 


993oi3 


• 38 


257269 12 


01 


742731 ! 45 


16 


250980 


n-61 


992990 


• 38 


257990 12 
258710 11 


00 


742010 


44 


H 


251677 


11-59 


992967 


• 38 


98 


741290 


43 


18 


252373 


n-58 


992944 


• 38 


259429 11 


96 


740571 


42 


*9 


253o67 


n-56 


992921 


• 38 


260146 11 


94 


739854 


41 


20 


253761 


u-54 


992898 


• 38 


260863 11 


92 


739137 


4r 


21 


9-254453 


11-52 


9.992875 


• 38 


9-261578 11 


.90 


10-738422 


U 


22 


255i44 


1,1.50 


992852 


• 38 


262292 11 


89 


737708 


23 


255834 


11.48 


992829 


•3 9 


263oo5 11 


ll 


736995 


37 


24 


256523 


11.46 


992806 


• 3 9 


263717 11 


736283 


36 


a5 


25721 1 


n-44 


992783 


• 3 9 


264428 11 


• 83 


735572 


35 


26 


257898 


11-42 


992759 


•3y 


265i38 11 


• 8i 


734862 


34 


27 


258583 


11-41 


992736 


.3 9 


265847 11 


' 7 2 


734i53 


33 


28 


259268 


ii-3 9 


992713 


-3 9 


266555 11 


•78 


733445 


32 


29 


259951 


11.37 


992690 


•3 9 


267261 11 


.76 


732 7 3o 


3i 


. 3o 


26o633 


11-35 


992666 


.39 


267967 11 


■74 


732033 


3o 


3i 


9-26i3i4 


n-33 


9-992643 


•3 9 


9-268671 11 


.72 


io-73i32o 


29 


*2 


261994 


n-3i 


992619 


•3 9 


269375 11 


■ 70 


730623 


28 


33 


262673 


n-3o 


992596 


•3 9 


270077 11 


.69 


729923 


27 


34 


263351 


11-28 


992572 


•3 9 


270779 11 


ii 


729221 


26 


35 


264027 


11-26 


992549 


•3 9 


271479 11 


728521 


25 


36 


264703 


11-24 


992520 


•3 9 


272178 11 


.64 


727822 


24 


3 7 


265377 


11-22 


992501 


•3 9 


272876 11 


■62 


727124 


23 


38 


266061 


11-20 


992478 


.40 


273573 1.1 


■60 


726427 


22 


3 9 


266723 


11-19 


992454 


40 


274269 11 


• 58 


725731 


21 


4o 


267395 


11-17 


992430 


•40 


274964 11 


•5 7 


725o36 


20 


4i 


9-268o65 


ii-i5 


9-992406 


•40 


9-275658 11 


.55 


10-724342 


IO 


42 


268734 


n-i3 


992382 


•40 


27635i 11 


• 53 


723649 


10 


43 


269402 


11 - 1 1 


992359 


•40 


277043 11 


• 5i 


722957 


17 


44 


270069 


1 1 • 10 


992333 


•40 


277734 11 


• 5o 


722266 


l6 


45 


270730 


u-o8 


9923n 


•40 


278424 11 


• 48 


721576 


i5 


46 


271400 


11-06 


992287 


•40 


279113 11 


•47 


720887 


14 


47 


272064 


n-o5 


992263 


•40 


279801 11 


•45 


720199 


i3 


48 


272->26 


n-o3 


992239 


•40 


280488 1 1 


• 43 


7 1 95 1 2 


12 


49 


273388 


II -01 


992314 


•40 


281174 11 


•4i 


" 718826 


11 


5o 


274049 


10-99 


992190 


•40 


281808 11 


•40 


718142 


10 


5i 


9-274708 


10.98 


9.992166 


.40 


9-282542 11 


• 38 


10-717458 


9 


52 


275367 


10-96 


992142 


•40 


283225 11 


• 36 


716775 


8 


53 


276024 


10-94 


992117 


•41 


283907 11 


.35 


7 1 6093 


7 


54 


276681 


10-92 


99 ^093 


•41 


284588 11 


• 33 


7 1 54 1 2 


6 


55 


277337 


10-OI 


992069 


•41 


285268 11 


• 3i 


7U732 


5 


56 


27799I 


10-89 


992044 


•41 


285947 11 


• 3o 


7i4o53 


4 


5 7 


278644 


10-87 


992020 


•41 


286624 n 


• 28 


713376 


3 


58 


279297 


10-86 


99,996 


•41 


287301 11 


• 26 


7 1 2699 
712023 


2 


5 9 


279Q48 


10.84 


991971 


•4l 


287977 11 


•25 


1 


60 


280399 


10 82 


991947 


•41 


288652 11 


•23 


711348 





Cosine 


D. 


Sine 


Cotanff. IJ 

B i 


1, 


Tang. 


M. 








(T9i 


)EOR 


EES.) 









\ 



BIKES AND TANGENTS. (11 DEGREES.) 



29 



o 
i 

2 

3 
4 
5 
6 

7 
8 

9 

10 

ii 

12 

i3 

U 
id 

16 

n 

18 

19 

20 

21 
22 
23 
24 
2D 
26 

27 

28 

29 
3o 

3i 

32 

33 

34 

35 

36 

3 

3 

39 
4o 

4i 
42 
43 
44 
45 
46 

47 
48 

49 
5o 

5i 

52 

53 

54 
55 
56 

57 
58 

59 

6o 



Sine 



D. 



Cosine D. 



Tang. 



D. Cotang. 



■ 2S0599 
281248 
281897 
282544 
2S3190 
283836 
284480 
235 124 
283766 
286408 
287048 

■287687 
288326 
288964 
289600 
290236 
290870 
291504 
292137 
292768 
293399 

1-294029 
294658 
295286 
295913 
296339 
297164 

' 297733 
298412 
299014 
299655 

5-300276 
300^93 
3oi5i4 

3o2l32 

302748 
3o3364 
3o397q 
3o4393 
303207 
3o58i9 

9-3o643o 
307041 
307650 
308259 
308867 
309474 
3 1 0080 
3io 

3112S9 
3 1 1893 

9-3i2495 
3 1 3097 
3 1 36q8 
314297 
314897 
3i5495 
316092 
316689 
317284 
317879 



10 


82 


10 


81 


10 


•79 


10 


•77 


10 


.76 



10-74 
10-72 
10-71 
10-69 
10-67 
io-66 

10-64 
io-63 

10-01 

io-5q 
io-53 
io-56 
io-54 
io-53 
io-5i 
10 -5o 

10-48 
10-46 
io-45 
io-43 
10-42 
10-40 
io-3g 
10-37 
10-36 
io-34 

10-32 

io-3i 



10 
10 



10-26 

10-25 
10-23 
10-22 
10-20 

io- 19 

10-17 

io- 16 
10- 14 
1 • 1 3 
io- 1 1 

10-10 

10-08 
10-07 
io-o5 
io-o4 



10 
10 
10 

9 
9 
9 
9 



o3 
01 
00 
9 3 

96 
. 94 

9- 9 3 
9-91 

9-9° 



9.991947 
991922 
991897 
991873 
991S4S 
091823 

091799 
991774 
991749 
991724 
99io99 

9.991674 
991649 
991624 
991599 
991574 
99i549 
99i524 
991498 
991473 
991448 

9.991422 
991397 
991372 
991346 
991321 
991293 
991270 

991244 
99 1 2 1 8 
991193 

9-991167 
991141 
9911 i5 

991090 
991064 
99io33 
991012 
990986 
990960 
990934 

9-990908 
990882 
99o355 
990829 
990803 

990777 
990750 

990724 
990697 
990671 

9-990644 
990618 
990591 
990565 
99o538 
990") 1 1 

99 
990 i 5 3 

990431 

9904 14 



•4i 
•41 
•4i 
•41 1 

t\ 

• 41 

•41 

•42 
.42 
•42 
•42 

•42 
.42 
.42 
•42 
•42 
•42 
•42 
•42 
•42 
•42 

.42 ; 

.42 

•43 

•43 

•43 

•43 

•43 

•43 

•43 

•43 

•43 
•43 
•43 
•43 
•43 
•43 
•43 
•43 
•43 
•44 

•44 

.AA 



•44 
•44 
•44 
•44 

•44 

•44 
•44 

•44 
•44 
•44 
•44 
•44 
•45 
•45 
■4> 
•45 
•45 



,-238652 
289326 

289999 

290671 

291342 
292013 
292682 
29335o 
294017 
294684 
295349 

9-296013 
296677 
297339 
298001 
298662 
299322 
299980 
3oo638 
301295 
301951 

9-302607 
3o326i 
303^14 
304367 
3o52i8 
3o5369 
3o65i9 
307168 
307815 
3o8463 

9.309109 
309754 
3io3g3 
3i 1042 
3u685 
312327 
312967 
3i36o8 

3U247 
3i4885 

•3i5523 
3 161 59 
316793 
3i743o 
3i8o64 
318697 
319329 
3 1 996 1 
320392 

321222 

? -32i85i 
322479 
323io6 
323733 
324353 
324Q83 
325607 
32623i 
3a6853 
327475 



11-23 
11-22 



II 

II 
II 
II 
II 
II 
II 
II 



20 
18 

H 

i5 
•14 

■12 
• II 
-09 



11-07 

11 -06 
11 -o4 
11 -o3 

II -01 
11-00 

10.98 

.0-96 
10-95 
10-93 
10-92 

10-90 
10-89 
10-87 
io-86 
10-84 
io-83 
io-8i 
io- 80 
10-78 
10-77 

10-75 
10-74 



10 
10 
10 
10 
10 



io- 

10 

10 

10 

10 

10 

10 

10 

10 

10 



10-711348 1 60 

I 7io°74 
710001 
709329 
708658 
707987 
707318 
7o665o 
705983 
7o53i6 
70465i 

10-703987 
7o3323 
702661 
701999 
70i333 
700678 
700020 
699362 
698705 
698049 



10 



io 



73 

7i 
70 

68 
.67 
io-65 
10-64 
10-62 

io-6i 

io- 60 

io-58 

io-57 

io-55 

io-54 

io-53 

io-5i 

10 -5o 

10-48 

47 
45 
44 
43 

41 

40 
3 9 

•36 

•35 



697393 
696739 
696086 
695433 
694782 
694131 
693481 
692832 
692185 
691537 

•690891 
690246 
689602 
688958 
6883 1 5 
687673 
687033 
686392 
685753 
685n5 

10-684477 
683841 
683205 
682570 
68m36 
68i3o3 
680671 
680039 
679408 
678778 

10-678149 
677521 
676894 
676267 
673642 
675017 
6 7 43 9 3 
673769 

673147 
672325 



37 
56 
55 
54 
53 

52 

5i 

5o 

ii 

40 

47 
46 
45 
44 
43 
42 
41 
40 

3 9 

38 

37 
36 

35 

34 
33 

32 

3i 
3o 

2 

27 
26 

25 

24 

23 
22 
21 

20 



Cosine 



D. 



Sine 



Cotang. 



I). 



Tan?. 



H 

H 
16 

i5 
14 
i3 
12 
11 
10 

9 
8 

7 
6 
5 
4 
3 
2 
1 
o 

M. 



(78 degrees) 



30 



(12 DEGREES.) A TABLE OF LOGARITHMIC 



M. ! 

-1 

o 


Sine 


D. 


Cosine 


D. 

45 


Tang. 


D. 


Cotang. 


60 


9-317879 


9 . 9 o 


9 • 990404 


9-327474 


io-35 


10-672526 


i 


318473 


9" 


88 


990378 


45 


328095 


10 


33 


671905 


5 9 


2 


3 1 9066 


9" 


87 


99o35i 


45 


328715 


10 


32 


671285 


58 


3 


3i 9 658 


9- 


86 


990324 


45 


329334 


10 


3o 


670666 


57 


4 


320249 


9' 


84 


990297 


45 


329953 


10 


29 


670047 


56 


5 


320840 


9 


83 


990270 


45 


3303TO 


10 


28 


669430 


55 


6 


3ai43o 


9 


82 


990243 


45 


331187 


10 


26 


6688 1 3 


54 


7 


322019 


9 


80 


990215 


45 


33i8o3 


10 


25 


668197 


53 


8 


322607 


9 


79 


990188 


45 


3324i8 


10 


24 


667562 


52 


9 


323i 9 4 


9 


77 


9901 61 


45 


333o33 


10 


23 


666067 


5i 


10 


323780 


9 


76 


990134 


45 


333646 


10 


21 


666354 


5o 


ii 


9 324366 


9 


75 


9-990107 


46 


9-334259 


10 


20 


10-665741 


49 


12 


324930 


9 


73 


990079 


46 


334871 


10 


19 


665129 


48 


i3 


325534 


9 


72 


990052 


46 


335482 


10 


17 


6645i8 


47 


U 


326117 


9 


70 


990025 


46 


336093 


10 


16 


663907 


46 


id 


326700 


9 


69 


989997 


46 


336702 


10 


i5 


6632 9 8 


45 


16 


327281 


9 


68 


989970 


46 


3373i 1 


10 


i3 


662689 


44 


17 


327862 


9 


66 


989942 


46 


337919 


10 


12 


662081 


43 


18 


328442 


9 


65 


989915 


46 


338527 


10 


11 


66i473 


42 


*9 


329021 


9 


64 


989887 
989860 


46 


339i33 


10 


10 


660867 


41 


20 


329D99 


9 


62 


46 


339739 


10 


08 


660261 


40 


21 


9.330176 


9 


61 


9-989832 


46 


9- 34o344 


10 


07 


10-659656 


39 


22 


33o753 


9 


60 


989804 


46 


340948 


10 


06 


659052 


38 


23 


33i329 


9 


58 


989777 


46 


34i552 


10 


04 


658448 


37 


24 


33 1903 


9 


5 7 


989749 


47 


342i55 


10 


o3 


657845 


36 


25 


332478 


9 


56 


989721 


47 


342757 


10 


02 


657243 


35 


26 


333o5i 


9 


54 


989693 


47 


343358 


10 


00 


656642 


34 


27 




9 


53 


989665 


47 


343958 


9 


.99 


656o42 


33 


28 


334195 


9 


52 


989637 


47 


344558 


9 


9 3 


655442 


32 


29 


334766 


9 


5o 


989609 


47 


345i57 


9 


97 


654843 


3i 


3o 


335337 


9 


49 


9 8 9 582 


47 


345755 


9 


96 


654245 


3o 


3i 


9'335906 


9 


48 


9-989553 


47 


9-346353 


9 


9i 


io-653647 


29 


32 


336475 


9 


46 


989525 


47 


346949 


9 


• 9 3 


653o5i 


28 


33 


337043 


9 


45 


989497 


47 


347543 


9 


•92 


652455 


27 


34 


3376lO 


9 


44 


989469 


47 


348141 


9 


91 


65i85 9 


26 


35 


338176 


9 


43 


989441 


47 


348735 


9 


9° 


65i263 


25 


36 


338742 


9 


41 


989413 


47 


349J29 


9 


• 88 


650671 


24 


3 7 


3393o6 


9 


40 


989384 


47 


349922 


9 


87 


650078 


23 


38 


339871 


9 


3 9 


9 8 9 356 


47 


35o5i4 


9 


• 86 


649486 


22 


3 9 


340434 


9 


37 


q8 9 328 


'47 


35 1 106 


9 


• 85 


648894 


21 


4o 


340996 


9 


36 


989300 


■47 


351697 


9 


• 83 


6483 o3 


20 


4i 


9-34i558 


9 


35 


9.989271 


•47 


9-352287 


9 


•82 


io-6477i3 


19 


42 


342119 


9 


34 


989243 


•47 


352876 


9 


■81 


647124 


18 


43 


342679 


9 


32 


989214 


■47 


353465 


9 


• 8o 


646535 


17 


44 


34323g 


9 


3i 


989186 


•47 


354o53 


9 


•79 


645947 


l6 


45 


343797 


9 


3o 


989157 


•47 


354640 


9 


77 


64536o 


i5 


46 


344355 


9 


29 


989128 


•48 


355227 


9 


.76 


644773 


i4 


47 


344912 


9 


27 


989100 


■48 


3558i3 


9 


■75 


644187 


i3 


48 


345469 


9 


26 


989071 


•48 


3563 9 8 
356982 


9 


■74 


643602 


12 


49 


346024 


9 


25 


989042 


•48 


9 


73 


643oi8 


11 


5o 


346579 


9 


24 


9S9014 


-48 


357566 


9 


'71 


642434 


10 


5i 


9-347i34 


9 


22 


9-988985 


• 48 


9-358i49 


9 


.70 


10641851 


9 


52 


347687 


9 


21 


9SS956 


.48 


35873i 


9 


• 69 


641269 


8 


53 


348240 


9 


■20 


988927 


•48 


3593i3 


9 


■ 68 


640687 


7 


54 


348792 


9 


19 


988898 


•48 


35 9 8 9 3 


9 


67 


640107 


6 


55 


349343 


9 


17 


988869 


-48 


360474 


9 


66 


639? j6 


5 


56 


349893 


9 


16 


988840 


•48 


36io53 


9 


65 


638 9 47 
638368 


4 


57 


35o443 


9 


i5 


98881 1 


•40 


36i632 


9 


63 


3 


58 


350992 
35 1 340 


9 


•14 


988782 


49 


362210 


9 


62 


637790 


2 


5 9 


9 


i3 


988753 


■49 


362787 


9 


61 


6372i3 


1 


6o 


352o88 


9. 11 


988724 


49 


363364 


9-60 


636636 


a 
M. 

.. . . — 


Com no 


D. 


Sine 




Cotang. 


D. 


Tang. 



(77 DEGREES.) 



SINES ASD TANGENTS. yW DEGREES.) 



31 



M.. 



Sine 



D. 



Cosine 



D. 



o 
i 

2 

3 

4 
5 
6 

7 
8 

9 

10 

ii 

12 

i3 

i5 

16 

18 

»9 

20 

21 
22 
23 

24 
23 

26 

27 
28 

11 

3i 

32 

33 

34 

35 

36 

3 

3 

3 9 

4o 



4i 
42 
43 

44 
45 
46 

4 

49 
5o 

5i 

52 

53 

54 

55 

56 

5 

5 

5o 

60 



9 . 352088 
352635 
353i8i 
353726 
354271 
3548i5 
355358 
355901 
356443 
356984 
307024 

9.358064 
3586o3 
359141 
359678 
36o2 1 5 
360752 
361287 
361822 
362356 
362889 

9-363422 
363954 
364485 
365oi6 
355546 
366075 
' 3666o4 
367131 
367659 
368i85 

9-368711 
36 9 236 
369761 
370285 
370808 
37i33o 
371852 
372373 
372894 
373414 

373933 
374452 

37497° 
375487 
376oo3 
376519 
377033 

37p49 
378063 

3 7 35 7 7 

,.379089 
379601 
38on3 
38o624 
38n34 
38i643 

3-$2l52 

382661 

383 1 68 
383675 



9. II 

9. 10 
9.09 
9.08 
9-07 
9-o5 
9.04 
9-o3 
9. 02 
9. 01 

8-99 
8-98 
8.97 
8-96 
8-g5 
8- 9 3 
8-92 
8-91 
8-90 
8-89 
8-88 

8-87 
8-85 
8-84 
8-83 
8-82 
8-8i 
8-8o 
8.79 
8.77 
8-76 

8- 7 5 
8-74 
8*^3 

8-72 
8.71 
8-70 
8-69 
8-67 
8-66 
8-65 

8-64 
8-63 
8-62 
8-6i 
8- 60 
8-5 9 
8-58 
8-57 
8-56 
8-54 

8-53 
8-52 
8-5i 
8-5o 
8-49 
8-48 

8-47 
8-46 
8-45 
8-44 



9-988724 
988695 
988666 
9 88636 
988607 
988578 
988548 
988519 
988489 
988460 
988430 

9.988401 
988371 
988342 
9 883 1 2 
988282 
988202 
988223 
988193 
988 1 63 
9 88i33 

9 - 9 88io3 
988073 
988043 
9880 1 3 
9 8 79 83 
987953 
987922 
987892 
987862 
987832 

9-987801 

9^777* 
987740 
987710 
987679 

987649 
987618 
9 8 7 588 
987557 
987526 

9-987496 
987463 
987434 
987403 

987341 
987310 
987279 
987248 
987217 

9-987186 
987 1 55 
987124 
987092 
987061 
987030 
986998 
986967 
986936 
986904 



Co3ine 



I). 



Tan*. 



D. 



Cotang. 



•49 
.49 
.49 
•49 
•49 
•49 

•49 
•49 
•49 
•49 
•49 

•49 
•49 

•49 

• 5o 

• 5o 

• 5o 
*5o 

• 5o 
•5o 

• 5o 

• 5o 
•5o 

• 5o 

• 5o 

• 5o 

• 5o 

• 5o 

• 5o 

• 5o 
-5i 

-5i 
*5i 

• 5i 

• 5i 

• 5i 

• 5i 

• 5i 

• 5i 

• 5i 

• 5i 

• 5i 
-5i 
-5i 

-52 
-52 
•52 
•52 
•52 
•52 
•52 

•52 
•52 
•52 
52 
•52 
•52 
•52 
■52 
•52 
•52 



Sine 



363364 
363940 
3645i5 
363090 
365664 
366237 
3663io 
367332 
367953 
368524 
369094 

•36 9 663 
370232 

370799 
371367 
371933 

372499 
373064 
373629 
374193 
374756 

•375319 
3 7 588i 
376442 
377003 
377563 
378122 
3 7 868i 
379289 

379797 
38o354 

1-380910 
38U66 

382020 

382575 

383i29 
333682 
384234 
384786 
385337 
335883 

? -386433 
386981 
3S 7 536 
388o34 
38863i 
SSgnS 
339724 
390270 
390815 
391360 

9-391903 

392447 
392989 

3 9 333i 
394073 
3946 1 4 
393154 
39369/, 
3962 5 3 
396771 



Cotang. 



9-60 
909 
9-58 
9-37 
9 -53 

9-54 
9-53 
9-52 
9-5i 
9-5o 
9-49 

9.48 
9-46 



40 
44 
43 
42 
9-41 
9-40 
9-3o 
9-33 

9.37 
9-35 

9-34 
9-33 
9-32 
9 • 3 1 

9-3o 

9.20 
9-28 
9-27 

9-26 
9- 25 
9-24 
9-23 
922 
9-21 
9-20 
9-ic 
9.18 
9-17 

9 • 1 5 
9-i4 
9- 13 
9- 12 
9. 11 
9- 10 
9.09 
9 - 08 
9-07 
9-06 

9-o5 
9 • 04 



o3 
02 
01 
00 

b£ 

8-97 
8-96 



10 



io*636636 

63bo6o 
635485 
634910 
634336 
633763 
633190 
682618 
632047 
631476 
630906 

io-63o337 
629768 
629201 
628633 
628067 
627501 
626936 
626371 
625807 
625244 

624681 
6241 10 
623558 
622997 
622487 
621878 
62 13 19 
620761 
620208 
619646 

•619090 

6i8534 
617080 
617425 
616871 
616818 
60766 

Si52i4 
614668 
614112 

io-6i3562 
6i3oi3 

612464 
61 1916 
611869 
610822 
610276 
609780 
609185 
608640 

10-60^097 
607553 
60701 1 
606469 
603927 
6o5 $86 
604846 
60 06 
608767 
603229 



10 



D. 



:*52& 



60 

5o 
58 

57 
56 
55 

54 
53 

52 

5i 
5o 

40 
48 

47 
46 
45 
44 
43 
42 
41 
40 

39 
38 

37 
36 
35 
34 
33 

32 

3i 

3o 

29 
28 

27 
26 

25 

24 

23 
22 
21 
20 

I 
It 

17 
16 

i5 
14 
i3 
12 
1 1 
10 



7 
6 

5 

4 

2 
1 
o 



27 



(7G DEGREES.) 



82 



(14 DEGREES.) A TABLE OF LOGARITHMIC 



M. 

o 

i 

2 

3 

4 
5 
6 

7 
8 

9 

10 

ii 

12 

i3 

U 
i5 
16 

«7 

18 

J 9 

20 

21 
22 
23 

24 

23 
26 

27 
28 

29 

3o 

3i 

32 

33 
34 
35 
36 

37 
38 

39 

4o 

4i 
42 
43 
44 
45 
46 

47 
48 

49 
5o 

5 1 
5a 
53 

54 
55 
56 
5 7 
58 
5 9 
6o 



Sine 



D. 



g. 383675 
384182 
384687 
385iq2 
380697 
386201 
386704 
387207 
387709 
388210 
388711 

9-389211 
3897 1 1 
390210 
390708 
391206 
391703 
392199 
3g26g5 
3g3 191 
3g36S5 

9.394179 
394673 
3g5i66 
395658 
396100 

396641 
397132 
397621 
398111 
398600 

9.399088 
399575 
400062 
400549 
4oio3o 
401 520 
402005 
402489 
402972 
4o3455 

9-403938 
404420 
404901 
4o5332 
4o5862 
4o634i 
406S20 
407299 

407777 
408254 

9-408731 
409207 
409682 
4ioi57 
4io632 
4 1 1 1 06 

4n579 

4I20D2 

412524 
412996 



8-44 
8-43 
8-42 
8-41 
8-40 
8-3 9 
8-38 
8-37 
8-36 
8-35 
3-34 

8-33 
8-32 
8-3i 
8-3o 
8-28 
8-27 
8-26 
8-25 
8-24 
8-23 

8-22 
8-21 
8-20 
8-IQ 

8-i8 
8-17 
8.17 
8-i6 
8-io 
8.14 

8- 13 

8-12 

8- 11 
8-io 

8-OQ 

8-o8 
8-07 
8-o6 
8-o5 
8-04 

8-o3 

8-02 

8-oi 
8-oo 
7.99 
7 - 9 S 

7-97 
7-96 

7-90 

7-94 

7-94 
7- 9 3 
7.92 
7.91 

7-9° 

7 -8§ 
7.87 
7-86 
7-85 



Cosine 



9-986904 
986873 
986841 
986809 
986778 
986746 
986714 
9 866S3 
986651 
986619 
986587 

9-986555 
9 36523 
986491 
986459 
986427 
986395 
9 86363 
9 8633i 
986299 
986266 

9-986234 
986202 
986169 
986137 
986104 
986072 
986039 
986007 
985974 
980942 

9-985909 
9 858 7 6 
985843 
9 858n 
985778 
985745 
985712 
985679 
985646 
9 856i3 

9-98558o 
985547 
9855i4 

985480 

935447 
985414 
9 8538o 

985347 
9853i4 
985280 

9-985247 
985213 
985180 
985146 
985i 1 3 
980079 
985o43 
98501 1 

984978 
984944 



D. 

-52 

.53 

• 53 

• 53 
-53 

• 53 

• 53 

• 53 

• 53 

• 53 

• 53 

-53 
.53 

• 53 

• 53 

• 53 
-53 
•54 
•54 
•54 
•54 

•54 
•54 
•54 
•54 
•54 
•54 
•54 
•54 
•54 
•54 

-55 
-55 

• 55 

• 55 
-55 

• 55 

• 55 

• 55 
.55 

55 

• 55 
■ 55 

• 55 

• 55 
-55 

• 56 
-56 

• 56 
-56 
-56 

• 56 

• 56 

• 56 

• 56 

• 56 

• 56 

• 56 

• 56 

• 56 

• 56 



Tang. 



D. 



Cosine 



D. 



Sine 



9.396771 
397309 
397846 
3 9 8383 
398919 
399455 
399990 
4oo524 
4oio58 
401591 
402124 

9-402656 
403187 
408718 
404249 
404778 
4o53oS 
4o5836 
4o6364 
406892 
407419 

9-407945 

408471 
408997 
409521 
410045 
410569 
411092 
4n6i5 
412137 
412658 

9-413179 
413699 
414219 
4U738 
4i5257 
4i5775 
416293 
416810 
417326 
417842 

9-4i8358 
418873 
419387 
4 1 990 1 
42041 5 
420927 
421440 
421952 
422463 
422974 

9-423484 
423993 
4245o3 
425ou 
4255i9 
426027 
426534 
427041 
427547 
428002 



Cutang. 



8-96 
8-96 
8-90 
8- 9 4 
8- 9 3 
8-92 
8.91 
8-90 
8*8 9 
8-83 
8-87 

8-86 
8-85 
8-84 
8-83 
8-82 
8-8i 
8-8o 
8-70 
8-76 
8-77 

8-76 

8-70 
8-74 
8-74 
8- 7 3 
8-72 
8.71 
8-70 
8-69 
8-68 

8-67 
8-66 
8-65 
8-64 
8-64 
8-63 
8-62 
8-6i 
8-6o 
8-5 9 

8-58 
8-57 
8-56 
8-55 
8-55 
8-54 
8-53 
8-52 
8-5i 
8-5o 

8.49 

8-48 
8-48 

8-47 
8-46 
8-45 
8-44 
8-43 
8-43 
8-42 



Cotang. 



10-603229 
602691 
602104 
601617 
6010S1 
6oo545 
600010 
099476 
• 598942 

- 598409 
597876 

10-597344 
5 9 68 1 3 
5962S2 
595751 
595222 
59469 i 
594164 
5 9 3636 
593108 
592081 

io-592o55 
591529 
591003 

590479 
589960 

58943 1 
588908 
588385 
587863 
587342 

10-586821 
5S63oi 
580781 
585262 
584743 
584225 
583707 
583190 
582674 
582108 

10-581642 
581127 
58o6i3 
580099 
579585 
079073 
57856o 
578048 
577537 
577026 

io-5765i6 

576007 
575497 
574989 
574481 
573973 
573466 
572909 
{172453 
1)71948 



P. 



60 

51 

57 
56 
55 

54 
53 

52 

5i 
5o 

40 

48 

47 
46 
45 
44 
43 
42 
41 
40 

3 9 

38 

37 
36 

35 

34 
33 

32 

3i 
3o 

29 

28 
27 
26 

25 

24 
23 
2? 
21 
20 

10 
10 

17 
16 

i5 
14 
-3 
. 2 

I ! 
10 



7 
6 
5 
4 
3 
2 
1 




Tang. M. 



(75 DEGREES.) 



SINES AND TANGENTS. (15 DEGEEES.) 



33 



M. 

o 

i 

2 

3 

k 
5 
6 



Sine 



D. 



Cosine D. Tang. 



D. 



Cotaug. 



9 

10 

ii 

12 

i3 
U 
i5 
16 

n 

18 
J 9 

20 

21 
22 
23 
24 
25 
26 

27 

28 

29 

3o 

3i 
32 
33 
34 
35 
36 

37 
38 

3 9 

40 

41 
42 
43 
44 
45 
46 

47 
48 

49 

DO 

5i 

5a 

53 

54 

55 

56 

5 

5< 

Sq 

60 



9.412996 
413467 
413938 
414408 
414878 

4i5347 
41 5si 5 
416283 
416751 
41721-' 
417684 

9-4i8i5o 
4i86i5 
419079 
419544 
420007 
420470 
4209.33 
421395 
421857 
4223i8 

9-422778 
423238 
423697 
4241 56 
4246i5 
425cr3 
42J53o 
425987 
426443 
426899 

9-427354 

427809 

42S263 

42 

429170 

429O23 

430075 

4'5o527 

43 

43U29 

9-43i8 79 
4 5 23 29 
432778 
433226 
4J36 7 5 
434122 
434569 
4 3 ')o 1 6 
43 3462 
435908 

9-4363 33 

436798 
437242 
437686 
438129 
438572 
439014 
439456 

439*97 
44o338 



7-85 

7-84 
7-83 
7-83 
7-82 
7-81 
7-80 

7-79 
7 - 7 8 

7-77 
7.76 



73 
74 
73 
73 
72 

7i 

70 

69 
7-68 
7.67 

7-67 
7 -60 
7-65 
7.64 
7-63 
7-62 
7-61 
7-60 
7-60 
7-5 9 

7-58 
7-57 
7-56 
7-55 
7-54 
7-53 
7-52 
7-52 
7 • 5i 
7-5o 

7-49 
7-49 

7-48 

7-47 
7.46 

7-45 
7-44 
7-44 
7-43 
7.42 



41 
■40 
.40 
-3o 

• 38 

•37 

• 36 

• 36 

• 35 
■34 



9.984944 ! 

9849IO ! 
984876 J 
984S42 
984808 

9^774 
984740 
984706 
984672 
9 8463 7 
984603 

9.984569 
984535 
984500 
984466 
984432 

984397 
9 84363 
984828 

984294 
984209 

9.084224 
984I9O 
984155 
984120 
984085 I 
984o5o i 
984015 
983981 i 
983946 j 
983911 I 

)-983875 j 
983840 

9 S33o5 
983770 
983735 
983700 
983664 
983629 
983594 
983553 

5-983523 
983487 
983452 
983416 
983381 
983345 
983309 
983273 
9 83238 
983202 

9- 9 83i66 
983i3o 
9S3oo.i 
q83o58 
9SJ022 
982986 
982950 

982914 

982878 



982842 



•57 
•57 
•57 
•57 
•57 
•57 
•5 7 
•57 
•57 
•57 
07 

•57 
•57 
•57 
•57 

• 58 

• 58 

• 58 

• 58 

• 58 

• 58 

• 58 

• 58 

• 58 

• 58 

• 58 

• 58 

• 58 

• 58 

• 58 

• 58 

• 58 

• 5 9 

• 5 9 

• 5 9 

• 5 9 

• 5 9 

• 5 9 
09 
09 

• 5 9 
09 
09 

5 9 
-5 9 

• 60 

• 60 

• 60 

• 60 

• 60 

• 60 

• 60 
.60 

• 6o 

• 6o 

• 6o 

• 6o 

• 60 



9.428052 
428557 
429062 
429566 
430070 
43o573 
431075 
43 1577 
432079 
43258o 
433o8o 

9.43358o 

/. 



4 34o3o 
434579 
435078 
435576 
436073 
436570 
437067 
437563 
438o59 

9-43S554 
439048 
439543 
44oo36 
440529 
441022 

44i5i4 
442006 

4424Q7 
442988 

9-443479 
443968 
444453 

444947 
445435 
445923 
446411 
446898 

447334 
447870 

9-448356 

448841 
449326 
449810 
450294 
450777 
43i26o 
451743 

452225 

452706 
9-453187 

453668 
454U3 
454628 
455io7 
455586 
456064 
45j542 
467019 
45749 6 



8.42 

8.41 

8.40 

8-3 9 

8-38 

8-38 

8-37 

8-36 

8-35 

8-34 

8-33 

8-32 
8-32 
8-3i 
8-3o 
8-29 
8-28 
8-28 
8-27 
8-26 
8-25 

8-24 
8-23 
8-23 

8-22 
8-21 
8-20 

8-19 
8-19 
8-18 
8.17 

8- 16 
8- 16 
8-i5 
8-i4 
8-i3 
8.12 

8-12 



11 

10 
09 



0-09 
8-o8 
8-07 
8-o6 
8-o6 
8-o5 
8-o4 
8-o3 

8-02 
8-02 

8-oi 
8- 00 

7-99 
7-9 

7-9 

7-97 

7-96 

7.96 

7- 9 j 
7-94 



10-571948 

571443 
570933 

570434 
569930 
569427 
568925 
568423 
567921 
567420 
566920 

10-566420 
665920 
565421 
564922 
564424 
563927 
56343o 
562933 
562437 
561941 

io-56i446 
560932 
56o457 
559964 
559471 
55-^978 
558436 
557994 
5575o3 
557012 

io-55652i 
556o32 
555 342 
555o53 
554565 
554077 
553JS9 
553 1 02 
5526i6 
552i3o 

55i644 
55 1 1 59 

55o6t4 

5 00190 
549706 

51Q2 23 
548740 
548237 

54--"3 
5472')4 

. 5 i68 1 I 
5463 1 2 
545852 

5 ;3>-2 
►893 

5 14 4 1 i 
Iq36 
543458 
542o8« 
543304 



10 



60 
5 9 
58 

57 
56 

55 
54 
53 

52 

5i 
5o 

49 
48 

47 
46 
45 
44 
43 
42 
4i 
40 

3 9 
38 

37 
36 

35 

34 
33 

32 

3i 
3o 

28 

27 
26 

25 
24 
23 
22 
21 
20 

19 
10 
17 
10 
1 5 
14 
i3 
12 

I i 
10 



Cosine 



D. 



Sine 



J?°idL n JiL__L _J 



D. 



Tang. 



1 
6 
5 
4 
3 
3 
1 




_VLj 



(74 DEGREES.) 



84 



(1G DEGREES.) A TABLE OF LOGARITHMIC 



M. 




Sine 


D. 


Cosine 


D. 


Tang. 


D. 


Cotnng. 


60 


9-44o338 


7-34 


9-982842 


• 6o 


9-457496 


7-94 


10 • 5425o4 


i 


440718 


7 


•33 


982805 


• 6o 


457973 


7 


- 9 3 


542027 


69 


2 


441218 


7 


•32 


982769 


•61 


458449 


7 


- 9 3 


54i55i 


58 


3 


44i658 


7 


•3i 


982733 


•61 


458925 


7 


•92 


541075 


57 


4 


442096 


7 


•3i 


9S2696 


•61 


459400 


7 


•91 


54o6oo 


56 


5 


442535 


7 


•3o 


982660 


•61 


459873 


7 


.90 


54oi25 


j 5d 


6 


442r//3 


7 


•29 


982624 


•61 


46o34o 


7 


.90 


539601 


54 


I 


443410 


7 


•28 


982587 


•6i 


460823 


7 


•89 


539177 


53 


44384T 


7 


•27 


982551 


•61 


461297 


7 


• 88 


5387o3 


52 


9 


444284 


7 


•27 


982514 


•61 


461770 


7 


• 88 


53823o 


5i 


10 


444720 


7 


•26 


982477 


•61 


462242 


7.87 


53 77 58 


5o 


ii 


9-445i55 


7 


•25 


9-982441 


•61 


9-462714 


7.86 


10-537286 


1 49 


12 


445590 


7 


•24 


982404 


•61 


463 1 86 


7 


• 85 


5368i4 


1 48 


i3 


446025 


7 


•23 


982367 


•61 


463658 


7 


■85 


536342 


47 


14 


4464D9 


7 


•23 


982331 


•61 


464129 


7 


•84 


535871 


46 


i5 


4463 9 3 


7 


•22 


982294 


•61 


464399 


7 


• 83 


5354oi 


45 


16 


447326 


7 


•21 


982207 


•61 


465069 


7 


■83 


53493i 


44 


H 


447759 


m 

1 


•20 


982220 


•62 


465539 


7 


• 82 


53446i 


43 


18 


448101 


1 


■20 


982183 


•62 


466008 


7 


• 81 


533992 
533524 


42 


l 9 


448623 


7 


•19 


982146 


.62 


466476 


7 


■ 8o 


4i 


20 


449054 


7 


•10 


982109 


•62 


466945 


7 


■ 80 


533o55 


4o 


21 


9-449485 


7 


•n 


9-982072 


•62 


9-467413 


7 


.79 


10-532587 


3 9 


22 


4499i5 


1 


•16 


982035 


.62 


467880 


7 


78 


532120 


38 


23 


45o345 


1 


.16 


981998 


•62 


468347 


7 


78 


53 1 653 


37 


24 


400775 


7 


•i5 


981961 


•62 


468814 


7 


77 


53 1 186 


36 


2D 


45i2o4 


7 


14 


981924 


•62 


469280 


7 


76 


530720 


35 


26 


45 1 632 


7 


i3 


981886 


•62 


469746 


7 


75 


53o254 


34 


27 


452o6o 


7 t3 


981849 


•62 


4702 1 1 


7 


75 . 


5*29789 


33 


28 


452488 


7-12 


981812 


•62 


470676 


7 


74 


52o324 


32 


29 


45290 


7 


11 


981774 


•62 


471141 


7 


73 


528859 


3i 


3o 


453342 


7 


i« 


981737 


•62 


471605 


7 


73 


528393 


3o 


3i 


9 -453 7 68 


7 


10 


9-981699 


•63 


9-472068 


7 


72 


10-527932 


29 


32 


454194 


7 


09 


981662 


•63 


472532 


7 


7i 


527468 


28 


33 


454619 


7 


08 


981625 


■63 


472995 


7 


7 1 


527005 


27 


34 


455o44 


7 


07 


981587 


•63 


473457 


7 


70 


526543 


26 


35 


455469 


7 


07 


981549 


•63 


473919 


7 


69 


526081 


25 


36 


455893 


7 


06 


981512 


•63 


47438i 


7" 


69 


5256i9 


24 


3 J 


4563i6 


7 


o5 


981474 


•63 


474842 


7- 


68 


525i58 


23 


38 


456739 


7 


04 


981436 


•63 


4753o3 


7- 


67 


524697 


22 


39 


457162 


7" 


04 


981399 


•63 


475763 


7- 


67 


524237 


21 


4o 


457584 


T 


o3 


- 981361 


•63 


476223 


1- 


66 


523 77 7 


20 


4i 


9-45Soo6 


V 


02 


9-981323 


•63 


9-476683 


1- 


65 


io.5233i7 


IO 


42 


458427 


T 


01 


981285 


•63 


477142 


7- 


65 


522858 


10 


43 


458848 


7* 


01 


981247 


•63 


477601 


7- 


64 


522399 


17 


44 


459268 


7- 


00 


981209 


•63 


478059 


7- 


63 


621941 


16 


45 


459688 


6- 


99 


981171 


•63 


4785i7 


7- 


63 


521483 


i5 


46 


460108 


6- 


9° 


98u33 


•64 1 


478975 


7" 


62 


521025 


14 


4^ 


460527 


6- 


98 


9810CP 


•64 


479432 


7- 


61 


52o568 


i3 


48 


460946 


6- 


97 


981037 


.64 


479889 


7- 


61 


5201 11 


12 


49 


46 1 364 


6- 


96 


981019 


•64 


48o343 


7- 


60 


5i9655 


11 


5o 


461782 


6- 


9 3 


980981 


•64 


480801 


7- 


5 9 


619199 


10 


5i 


9-462199 


6- 


9 5 


9-980942 


.64 


9-481257 


7- 


5 9 


ic-5i8743 


i 


52 


462616 


6- 


9 4 


980904 


.64 


481712 


7- 


58 


518288 


53 


463 o3 2 


6- 


9 3 


980866 


•64 


482167 


7- 


57 


5i7833 


7 


54 


463448 


6- 


9 3 


980827 


•64 


482621 


7- 


57 


517379 


6 


55 


463864 


6- 


92 


980789 


•64 


483075 


7- 


56 


516923 


5 


56 


464279 


6- 


9i 


980750 


•64 


483529 


7" 


55 


616471 


4 


57 


464694 


6- 


90 


980712 


.64 


483982 


7- 


55 


5i6oi8 


3 


58 


465 108 


6- 


9° 


980673 


.64 


484435 


1 ■ 


54 


5i5565 


2 


5 9 


465522 


6- 


89 


980635 


•64 


484887 


7- 


53 


5i5n3 , 


1 


6o 


465935 


6-88 


980596 


•64 

i 


485339 


7-53 


5 1 466 1 ! 

_ j 





Cosine 


D. 


Sine 1 


Cotauer. 


D. 


Tang. 


ML 



(73 DECREES.) 



SINES AND TANGENTS. (17 DEGREES.) 



35 



M. 



Sine 



D. 



Cosine D. 



o 
I 

i 
3 
4 
5 
6 

7 
8 

9 

10 

ii 
12 

i3 

U 
i5 

16 

17 
18 

19 
20 

21 
22 

23 

24 

25 

26 

2 
2 

12 

3i 

32 

33 

34 

35 

36 

3 

3 

3 9 

40 

41 
42 
43 
44 
45 
46 

47 
48 

it 

5i 

52 

53 
54 
55 
56 

tl 



•463935 
466348 
466761 
467173 
467535 
467996 
468407 
468817 

469227 
469637 
470046 

•470455 
470863 
471271 
471679 
472086 

472492 
472898 

473304 
473710 
474i i5 

!-4745io 
47492J 
475327 
475730 
476i33 
476536 
476938 
477340 
477741 
478142 

)■ 478542 
478942 
479342 
479741 
480 1 40 
480539 
48o 9 3 7 
43 1 334 
481731 
482128 

5-482525 
482921 
4833 1 6 
483712 
484107 
4845oi 
434895 
43523 9 
485682 
486075 

9-486467 
486860 
487251 
4876;] 
488034 
488474 
488814 
489204 
489393 
489982 



Cosine 



6-88 
6-88 
6-87 
6-86 
6-85 
6-85 
0-84 
6-83 
6-83 
6-8f 
6-8[ 

6-8o 
6-83 
6-7Q 

6.78 
6-78 

6-77 
6-76 
6-76 
6-75 
6-74 

6-74 
6-73 
6-72 
6-72 

6.71 
6-70 
6-69 
6-6 9 
6-68 
6-67 

6-67 
6-66 
6-65 
6-65 
6-64 
6-63 
6-63 
6-62 
6-6i 
6-6i 

6-6o 
6 09 
6-5 9 
6-58 
607 
6- 5 7 
6-56 
6-55 

6-33 

6-54 



53 
53 

52 

5i 

5i 
6-5o 
6-5o 

6-49 
6-43 
6.48 



9-980596 
980358 
980519 
980480 
980442 
980403 
9 8o364 
98o325 
980286 

980247 
980208 

9-980169 
980130 
980001 
980032 
980012 
979973 
979934 
979895 
979355 
979816 

9.979776 

979737 
979697 
979638 
979618 
979579 
979539 

979499 
979439 
979420 

9-979380 
979340 
979300 
979260 
979220 
979180 
979140 
979100 
979059 
979019 

9.978979 
978939 
978898 
978838 
978817 

978777 
978736 
978606 
978635 
978615 

>• 978374 
978533 
078493 
978432 
978411 
978370 
978329 
978288 

978247 
978206 



Tang. 



D. 



Cotang. 



-64 
.64 
■65 
•C5 
•65 
•65 
•65 
•65 
•65 
•65 
•65 

• 65 

• 65 

• 65 

• 65 

• 65 

• 65 

• 66 

• 66 

• 66 

• 66 

■ 66 

• 66 

• 66 

• 66 

• 66 

• 66 

• 66 

• 66 

• 66 
•66 

• 66 
•66 
.67 
.67 
.67 
.67 
.67 
.67 
.67 
.67 

.67 
.67 
.67 
.67 

•67 
.67 

•67 

• 68 
•63 
•63 

• 63 
-68 

• 68 

• 68 

• 68 

• 63 

• 63 

• 68 

• 68 

• 68 



9-485339 
485791 
486242 
486693 
4S7143 
487593 
488043 
488492 
488941 
489390 
489333 

9-490286 
490733 
49 1 1 80 
491627 
492073 
492519 
492965 
493410 
493854 
494299 

9-494743 
495i86 
4g563o 
496073 
4965i5 
496957 
497399 
497841 
498282 
498722 

•499163 
499603 
5ooo42 
5oo48i 
500920 
5oi359 
50170] 

502233 
502672 

5o3io9 

>-5o3546 
5o3 9 32 
5o44i3 
5o4854 
503289 
5o5724 
5o6i5o 
506593 
507027 
507460 

>-5o7893 
5o8326 
508759 
509191 
509622 
5 1 oo54 
5 1 0485 
5 1 09 1 6 
5n346 
5i 1776 



7-33 
7-52 
7 .5i 
7 .5i 

7-5o 

7-49 
7-49 
7-43 
7-47 
7-47 
7.46 

7-^6 
7.45 
7-44 
7-44 
7.43 
7-43 
7.42 

7-4i 
7.40 
7.40 



40 
3o 
38 

37 
37 
36 
7-36 
7-35 
7-34 
7-34 

7-33 
7-33 
7.32 
7-3i 
7.3i 
7-3o 
7-3o 
7.29 

7-28 

7.28 



27 
27 
26 

25 
25 
24 
24 
23 
■22 
■22 



7-21 

7.21 
7-20 

7-19 
7.19 

7- 18 

7. 18 
7-n 

7- 16 
7- 16 



io-5i466i 
514209 
5i3758 
5i33o7 
5i2857 
512407 
5i 1937 
5ii5o8 
5uo59 
5 1 06 10 
5ioi62 

10-509714 
509267 
5o882o 
5o33 7 3 
507927 
507481 
5o7o35 
506390 
5o6i46 
5o57oi 

io-5o5257 

5o48i4 
504370 
5o3q27 
5o3485 
5o3o43 
5o26oi 
5o2i5o 
501718 
501278 

iO'5oo837 
5oo3o7 
499938 
499319 
499080 
498641 
498203 
497765 
497328 
496891 

10-496434 
4960 1 8 
495582 
495146 

4947 1 1 
494276 

493841 
493407 
492973 
492640 

10-492107 
49 1 674 
49I24I 
A 1 ) 

490 378 
jo46 
489 ii 5 
4S00H4 
488654 

488224 



D. 



Sine 



D. I Cotang. 



D. 



Tai.L 



60 

57 
56 

55 
54 
53 

52 

5i 

DO 

49 
48 

47 
46 
45 
44 
43 
42 
41 
40 

39 

38 

37 
36 

35 

34 
33 

32 

3i 
3o 

29 
28 

27 
26 

25 

24 

23 
22 
21 
20 

IO 
10 

17 

16 
i5 

14 
i3 
12 
11 
10 

I 



5 
4 
3 
2 
1 




(72 DEGREES.) 



36 



(18 DEGREES.) A TABLE OF LOGARITHMIC 



M. 
o 


Sine 


D. 


Cosine 


D. 


Tan-. 


D. 


Cotang. 




9-489982 


6.48 


9-978206 


•68 


9-511776 


7.16 


10-488224 


60 


I 


490371 


6-48 


978165 


•68 


512206 


7 


.16 


487794 


So 


2 


490739 


6-47 


978124 


• 68 


512635 


7 


• ID 


487365 


58 


3 


491 147 


6-46 


978083 


.69 


5i3o64 


7 


•14 


486q36 


57 


4 


49i535 


6.46 


978042 


.69 


5i3493 


7 


•U 


. 486507 


56 


5 


491922 


6-45 


978001 


.69 


513921 


7 


-i3 


486079 


55 


6 


492308 


6-44 


977939 


.69 


5i4349 


7 


• i3 


48565i 


54 


I 


492695 


6-44 


977918 


•69 


514777 


7 


■ 12 . 


485223 


53 


493081 


6-43 


977877 


•69 


5i52o4 


7 


• 12 


484796 


52 


9 


493466 


6-42 


977835 


.69 


5i563i 


7 


-ii 


484369 
483 9 43 


5i 


lo 


49385i 


6-42 


977794 


.69 


5i6o57 


7 


10 


5o 


ii 


9-4g4236 


6-41 


9-977732 


-69 


9-516484 


7 


■10 


io-4835i6 


4 2 


12 


494621 


6.41 


9777" 


.69 


516910 


7 


09 


483090 


48 


i3 


490005 


6-40 


977669 


.69 


517335 


7 


■ 09 


482665 


47 


14 


49^388 


6-3 9 


977628 


.69 


517761 


7 


■ 08 


482239 


46 


i5 


493772 


6-3 9 


977386 


.69 


5i8i85 


7 


.08 


481815 


45 


16 


496154 


6-38 


977544 


.70 


5i86io 


7 


■ 07 


481390 


44 


17 


496537 


6-3 7 


977303 


•70 


519034 


7 


■ 06 


480966 


43 


18 


496919 


1 6-3 7 


97746i 


.70 


5i9458 


7 


■ 06 


480342 


42 


*9 


4973oi 


6-36 


9774i9 


•70 


519882 


7 


•o5 


4801 18 


4i 


20 


497682 


6-36 


977377 


.70 


52o3o5 


7 


■03 


479695 


40 


21 


9-498064 


6-35 


9-977335 


.70 


9-520728 


7 


.04 


10-470272 


3 9 


22 


498444 


6-34 


977293 


.70 


52i i5i 


7 


•o3 


478849 


38 


23 


498825 


6-34 


977231 


.70 


521573 


7 


o3 


478427 


b 


24 


490204 


6-33 


977209 


• 70 


521995 


7 


■o3 


478005 


36 


25 


499584 


6-32 


977167 


.70 


522417 


7 


■C2 


477583 


35 


26 


499963 


6-32 


977125 


.70 


522838 


7 


■02 


477162 


34 


27 


5oo342 


6-3i 


977083 


.70 


523259 


7 


■01 


476741 


33 


28 


500721 


6-3i 


977041 


•70 


52368o 


7 


01 


476320 


32 


29. 


501099 


6-3o 


976999 


.70 


524100 


7 


■00 


475900 


3i 


3o 


501476 


6-29 


976937 


• 70 


524520 


6 


99 


475480 


3o 


3i 


9-5oi854 


6-29 


9.976914 


.70 


9-524939 


6 


°9 


io-475o6i 


20 


32 


5o223l 


6.28 


976872 


•7i 


523359 


6 


98 


474641 


28 


33 


502607 


6-28 


97683o 


•71 


525 77 S 


6 


98 


474222 


27 


34 


502984 


6-27 


976787 


•7i 


526197 


6 


97 


4738o3 


26 


35 


5o336o 


6-26 


976745 


•7i 


5266i5 


6 


97 


4733S5 


25 


36 


5o3735 


6-26 


976702 


•7i 


527o33 


6 


96 


472967 


24 


37 


5o4uo 


6-25 


976660 


•7i 


527451 


6 


96 


472349 


23 


38 


5o4485 


6-25 


976617 


.71 


527868 


6 


9 5 


472i32 


22 


3 9 


5o486o 


6-24 


97 65 7 4 


•7i 


528285 


6 


9 5 


471715 


21 


4o 


5o5234 


6-23 


976532 


•7i 


528702 


6 


94 


471298 


20 


41 


9-5o56o8 


6-23 


9-976489 


•7i 


9-529119 


6 


9 3 


10-470881 


IO 


42 


5o5 9 8i 


6-22 


976446 


•7i 


529333 


6 


93 


470465 


10 


43 


5o6354 


6-22 


976404 


•7i 


529950 


6 


93 


470o5o 


n 


44 


506727 


6-21 


976I61 


•7i 


53o366 


6 


92 


469634 


16 


45 


507099 


6-20 


676318 


•7i 


5307S1 


6 


9 1 


469219 


i5 


46 


507471 


6-20 


976275 


•7i 


531196 


6 


9 1 


468804 


14 


47 


507843 


6-19 


976232 


.72 


53i6u 


6 


90 


468389 


13 


48 


5o82i4 


6-19 


976189 


.72 


532025 


6 


90 


467973 


12 


49 


5o8585 


6-i8 


976146 


•72 


532439 


6 


89 


467561 


II 


5o 


5o8 9 56 


6-i8 


976103 


.72 


532853 


6 


89 


467147 


13 


5i 


9.509326 


6.17 


9-976060 


•72 


9-533266 


6 


88 


10-466734 


Q 


5s 


509696 


6-i6 


976017 


•72 


533679 


6 


88 


466321 


6 


53 


5ioo65 


6- 16 


975974 


•72 


534092 


6 


87 


465908 


I 


54 


5io434 


6-i5 


975930 


•72 


5345o4 


6 


87 


4654o6 


55 


5io8o3 


6-i5 


975887 


.72 


534916 


6- 


86 


465o84 


5 


56 


5i 1 172 


6-U 


975844 


•72 


535328 


6- 


86 


464672 


4 


5 7 


5ii54o 


6i3 


975800 


•72 


535739 


6- 


85 


464261 


3 


58 


5i 1907 


6-i3 


975757 


•72 


536 1 5o 


6- 


85 


46385o 


2 


5 9 


512275 


6-12 


975714 


.72 


53656i 


6 


84 


46343o 


1 


60 


512642 


6-12 


975670 


•72 


536972 


6-84 


463028 



M. 




Cosine 


D. 


Sine 


D. 


Cotan?. 


D. 


Tan*. 



(71 DEGREES.) 



SINES AND TANGENTS. (19 DEGREES.) 



37 



M. 



Sine 



o 

i 

2 

3 

4 
5 
6 

7 
8 

9 

10 

n 

12 

i3 

U 

i5 

16 
17 
10 

19 

20 

21 

22 

23 

24 

25 

26 

2 1 
28 

29 

3o 
3i 

32 

33 

34 

35 

36 

3 

3 

39 
40 



41 

42 
43 

44 

45 

46 

47 
48 

49 
5o 

5i 

52 

53 

54 

55 

56 

5 

5 

59 

60 



D. 



^ 1 



). 512642 
5 i 3009 
5i337D 
5i374i 
514107 
5i4472 
5U837 

5ID202 

5 i 5566 
5i5g3o 
516294 

5.516657 
517020 
517382 
5i7745 
518107 
518468 
518829 
519190 
5iq55i 
5 1 99 1 1 

9.520271 
52o63i 
520990 
52i349 
D21707 
522066 

522424 
522781 
523 138 
523495 

g.523852 
524208 
524564 
524920 
525275 
52563o 
523984 
526339 
526693 
527046 

9.527400 
527753 
528io5 
528458 
528810 
629161 
5295i3 
529864 
53o2i5 
53o565 



9 .5309 1 5 
53i265 
53i6i4 
53ia63 
5323i2 
532661 
533oo9 
533357 
533704 
524o52 



6-12 

6- 11 

6-u 
6-io 



Cosine 



D. 



Tan.?. 



09 
09 



6-o8 



08 

°7 
07 
06 



6-o5 
6-o5 
6-04 
6-04 



o3 
o3 
02 
01 
6-oi 
6-oo 



L_J 



Cosine 



6-oo 
5-99 
5-99 
5-98 
5.98 
5-97 
5- 9 6 
5.96 
5.95 
5- 9 5 

5-94 
5-94 
5- 9 3 
5- 9 3 
5-92 
5.91 
5.91 
5.90 
5.90 
5.S9 

5-89 
5-88 
5-88 

5-87 
5-87 
5-86 
5-86 
5-85 
5-85 
0-84 

5.84 
5-83 
5-82 
5-82 
5-8i 
5-8i 
5-8o 
5-8o 

5' 7 2 
5.78 



9.975670 
973627 
975583 
975539 
975496 
970402 
97 54o3 I 
975365 J 
975321 
975277 
970233 

9.975189 
975142 
975101 
975007 
975oi3 
974969 
97492a 
974880 
974336 

97479 2 

9.974748 
974708 
974609 

9746U 
974570 
974525 
974481 
974436 
974391 
974347 
? - 9 743o2 
974257 
974212 
974167 
974122 

974077 
9i4o32 
973987 
973942 
973897 

9-973852 
973807 
973761 
973716 
973671 
973625 
973080 
973535 
973489 
973444 

9-973398 
973302 
973307 
973261 
9732i5 
973169 
973i24 
973078 
973o32 
972986 



D. 



73 
73 
73 
73 
73 
73 
73 
73 
73 
73 
73 

73 
73 
73 
73 
73 



74 



74 



D. 



Cotang. 



73 
75 
75 
75 
75 
73 
73 
73 
75 
7 5 

7 5 
76 

76 

76 

76 

76 

76 

76 

76 
76 
7 6 
76 
76 

76 
76 
76 

77 
77 



9.536972 
537382 
537792 
533202 
5386n 
539020 
539429 
53 9 337 
540245 
54o653 
541061 

9-54U63 

541875 

542281 

542688 

543094 

543499 
543905 

5443io 
5447i5 
545119 

9.545524 
545928 
54633i 
546735 
547i38 
547540 
547943 
548345 
548747 
549U9 

9.549550 

549901 
5oo3o2 
55o752 
55i 1 52 
55i552 
551952 
5o235i 
55275o 
553i49 

9.553548 
553o46 
554344 
554741 
5 5 5 1 3 9 
555536 
555o33 
5 563 2 9 
556720 
557121 

•557517 
557913 
5583o8 
508702 
509097 

55940' 

559885 

560279 
56o6 7 5 
56 1 066 



Sine 



D. 



Cotang. 



6-84 


6-83 


6-83 


6-82 


6-82 


6-8i 


6-8i 


6-8o 


6-8o 


6-79 


6-79 


6-78 


6-78 


6.77 


6-77 


6-76 


6-76 


6-75 


6-75 


6-74 


6-74 



110-463028 

462618 
462208 

461798 
461389 
460980 
460071 
460163 
459755 

439347 

408939 

■458532 
458i25 

407719 
457312 

406906 

456ooi 

406095 

455690 

455285 

454881 



10 



6-73 
6-',3 
6-72 
6-72 
6-71 

6-7i 
6-70 
6-70 
6-69 
6-69 

6-68 
6-63 
6-67 
6-67 
6-66 
6-66 
6-65 
6-65 
6-65 
6-64 

6-64 
6-63 
6-63 
6-62 
6-62 
6-6i 
6-6i 
6-6o 
6-6o 
6-59 

6-59 
6-5 9 
6-58 
6-58 
6-57 
607 
6-56 
6-56 
6-55 
6-55 



10 



10-454476 
454072 
45366a 
4532&0 
452862 
452460 
452007 
45 1 655 
45i253 
45o85i 

•45o45o 
45oo49 
449648 
449248 
443848 
448448 
448048 

447649 
44725o 
44685i 

10-446452 
446o54 
445656 
4452? 9 



io-4 



444464 



D. 



444861 
i4464 
$4067 
443671 
443270 

442879 

12483 
442087 

441692 
4 \ 1 298 
44090 J 
440O09 
4401 1 
439721 
439327 
438934 



60 

58 ! 

57 

56 

55 
54 
53 

52 

5i 
5o 

49 

48 

47 
46 
40 
44 
43 
42 
4i 
40 

39 
38 

37 
36 
35 
34 
33 

32 

3i 
3o 

2 2 
28 

27 

26 

25 

24 

23 
22 
21 
20 

% 

17 
16 

i5 
14 
i3 
12 
11 
10 



Tang. 



7 
6 
5 
4 
3 
2 
1 
o 

M. 



(70 DEGREES.) 



88 



(20 DEGREES.) A TABLE OF LOGARITHMIC 



M. 




Siae 


D. 


Cosine 


D. 


Tang. 


D. 


Cotang. 


' 60 


9-534o52 


5-78 


9-972986 


•77 


I 

I 9-561066 


6-55 


10-438934 


i 


534399 


5-77 


972940 


'77 


561459 


6-54 


438341 


1 59 


2 


534745 


5-77 


972894 


•77 


56i85i 


6-54 


438149 


58 


3 


530092 


5-77 


| 972848 


•77 


562244 


6-53 


437756 


57 


4 


535433 


5-76 


972802 


•77 


562636 


6-53 


437364 


56 


5 


535 7 83 


5-76 


972755 


•77 


563028 


6-53 


436C72 


55 


6 


536129 


5-75 


972709 


•77 


563419 


6-52 


43638i 


54 


7 


536474 


5-74 


972663 


•77 


5638n 


6-32 


436189 


53 


8 


5368i8 


5-74 


972617 


•77 


564202 


6-5i ■ 


435798 


52 




537i63 


5-73 


972570 


•77 


564592 


6-5i 


4354o8 


5i 


io 


537507 


5-73 


972524 


•77 


564933 


6-5o 


435oi7 


5o 


II 


9-53785i 


5-72 


9-972478 


•77 


9-565373 


6-5o 


10-434627 


4 2 


12 


538i 9 4 


5-72 


97243i 


.78 


565763 


6-49 


434237 


48 


i3 


533538 


5.71 


972385 


.78 


566 1 53 


6-49 


433847 


47 


U 


53888o 


5-71 


972338 ' 


.78 


566542 


6-49 


433458 


46 


ID 


539223 


5-70 


972291 


.78 


566932 


6-48 


433o68 


43 


16 


539565 


5-70 


972245 


.78 


567320 


6-48 


43268o 


44 


17 


539907 


5-6 9 


972198 


.78 


567709 


6-47 


432291 


43 


18 


540249 


5-6 9 


972151 


.78 


568098 


6-47 


43ioo2 


42 


*9 


540590 


5-68 


972io5 


.78 


568436 


6-46 


43iOU 


4i 


20 


540931 


5-68 


972o58 


.78 


568873 


6-46 


431127 


40 


21 


9-541272 


5-67 


9-972011 


.78 


9-569261 


6-45 


10-430739 


3 9 


22 


54i6i3 


5-67 


97 1 964 


.78 


569648 


6-45 


43o352 


38 


23 


541953 


5-66 


971917 


• l8 « 


570035 


6-45 


429965 


37 


24 


542293 


5-66 


971870 


.78 


570422 


6-44 


429078 


36 


23 


542632 


5-65 


971823 


.78 


570809 


6-44 


429191 


35 


26 


542971 


5-65 


971776 


.78 


571193 


6-43 


428800 


34 


27 


5433 10 


5-64 


971729 


•79 


571081 


6-43 


42S419 


33 


28 


543649 


5-64 


971682 


•79 


571967 


6-42 


428o33 


32 


29 


543987 


5-63 


971635 


•79 


572352 


6-42 


427648 


3; 


3o 


544325 


5-63 


971088 


•79 


572738 


6-42 


427262 


3o 


3i 


9-544663 


5-62 


9-971540 


•79 


9-573i23 


6-41 


10-426877 


29 


32 


545ooo 


5-62 


971493 


•79 


573507 


6-41 


426493 


28 


33 


545338 


5-6i 


971446 


•79 


573892 


6-40 


426108 


27 


34 


545674 


5-6i 


971398 


•79 


574276 


6-40 


425724 


26 


35 


54601 1 


5-6o 


971331 


•79 


574660 


6-39 


425340 


25 


36 


546347 


5-6o 


97i3o3 


•79 


575o44 


6-3 9 


424906 


24 


37 


546683 


5.59 


971256 


•79 


575427 


6-39 


424573 


23 


38 


547019 


5-59 


971208 


•79 


570810 


6-38 


424190 


23 


3 9 


547354 


5-58 


971161 


•79 


576193 


6-38 


423807 


21 


4o 


547689 


5-58 


97Hi3 


•79 


576576 


6-37 


423424 


20 


4i 


9.548024 


5-57 


9-971066 


• 8o 


9-576953 


tn 


io-423o4i 


IQ 


42 


54835 9 


507 


971018 


.80 


577341 


422659 


18 


43 


548693 


5-56 


970970 


• 80 


577723 


6-36 


422277 


1 1 


44 


549027 


5-56 


970922 


.80 


578104 


6-36 


421896 


16 


45 


549360 


5-55 


970874 


.80 


578486 


6-35 


42i5i4 


i5 


46 


549693 


5-55 


970827 


.80 


57S867 


6-35 


42ii33 


14 


47 


55oo26 


5-54 


970779 


.80 


579248 


6-34 


420702 


i3 


48 


55o359 


5-54 


970731 


• 8o 


579629 


6-34 


420371 


12 


49 


550692 


5-53 


970683 


• 8o 


580009 


6-34 


419991 


11 


5o 


55io24 


5-53 


970635 


.80 


58o3S 9 


6-33 


419611 


10 


5i 


9-55i356 


5-52 


9-970586 


.80 


9-580769 


6-33 


10-419231 

4i885i 





52 


55i68 7 


5-52 


970538 


.80 


581149 


6-32 





53 


552018 


5-52 


970490 


.80 


58i528 


6-32 


418472 


7 


54 


552349 


5-5i 


970442 


.80 


581907 


6-32 


418093 


6 


55 


55268o 


5-5i 


970394 


• 8o 


582286 


6-3i 


4i77 J 4 


5 


56 


553oio 


5-5o 


970345 


• 8i 


582665 


6-3i 


417335 


4 


tl 


553341 


5-5o 


970297 


.81 


583o43 


6-3o 


416957 


3 


553670 


5-49 


970249 


.81 


58342 2 


6-3o 


416578 


2 


59 


554000 


5-49 


970200 


•81 


5838oo 


6-29 


416200 


1 


60 

1 


554329 


5-48 


970i52 


• 8i 


584177 


6-29 


4i5823 




M. 


Coeine 


D. 


Sine 


D. 


Cotang. 


D. 


Tang. 



(69 DEGREES.) 



SIXES AND TANGENTS. (21 DEGREES.) 



30 



M. 



Sine 



D. 



Cosine 



D. 



Tang. 



o 

i 

2 

3 

4 
5 
6 

7 
8 

9 

10 

ii 

12 

i3 
14 

i5 
16 

\l 

19 
20 

21 

22 

23 

24 

23 
26 
2 
2 

II 

3i 

32 

33 

34 
3a 
36 
3 7 
38 
3 9 
40 

41 
42 
43 
44 
45 
46 

47 
48 

8 

5i 
5a 
53 
54 
55 
56 

ll 
59 
60 



•554329 

55-+658 
5549S7 
5553 1 5 
555643 

535971 
556299 
556626 
556953 
557280 
557606 

•537932 
558258 

553583 
558909 

5592.H 
559558 



560207 
56o53i 
56o855 

9061178 
56i5oi 
561824 
562146 
562468 
562790 
563 1 1 2 
563433 
563-55 
564075 

9-564396 
564716 
065o36 
565356 
565676 
565995 
5663 1 4 
566632 
566951 
567269 

9.567:587 
367;/) \ 
568222 
56853 9 
568856 
569172 
5j&q 

569804 
570120 
570435 

570751 

571066 

5713 lo 

571695 

572009 

572823 

5 7 2636 

5729)0 

573263 

573375 



9 



5-48 
5-48 
5-47 
5-47 
5-46 
5-46 
5-43 
5-43 
5-44 
5-44 
5-43 



43 

43 
42 
42 
41 
41 
40 
5-4o 
5-39 
5-39 

5-38 
5-38 
5-37 
5-37 
5-36 
5-36 
5-36 
5-35 
5-35 
5-34 

5-34 
5-33 
5-33 
5-32 



32 
3i 
3i 
3i 
3o 



5-3o 

5-29 
6-29 
5-28 
5-28 
3-28 
6-27 
5-27 
5-26 
5.26 
5-25 

5-25 
5-24 
5-24 
5-23 
5-23 
5-23 

5-22 
5«22 
5-21 
5-21 



9-970152 
970103 
970055 
970006 
969957 
969909 
969860 
96981 1 
969762 

969714 
969665 

9-969616 
969567 
969518 
969469 
969420 
969370 
969321 
969272 
969223 
969173 

9-969124 

969075 
969023 
968976 
968926 
968877 
968827 
96877-7 
968728 
968678 

9-968628 

9 685 7 8 
9 68528 

968479 
968429 
968379 
968329 
968278 
968228 
968178 

9-968128 
968078 
968027 
967977 
967927 
967870 
967826 

967.T75 

967725 

Q67674 

9-967624 
967573 
9 / >752 2 

967471 
967421 
967370 
967319 
967268 
9672:7 
967166 



Cosine 



D. 



.81 
.81 
.81 

• 81 
•8i 
.81 
•8i 
•81 
.81 

• 8i 

• 81 

.82 
•82 
.82 
.82 
.82 
.82 
•82 
•82 
.82 
•82 

.82 
.82 
.82 
.82 

• 83 

• 83 
.83 

• 83 
.83 

• 83 

• 83 

• 83 
•83 
-83 

• 83 

• 83 

• 83 

• 83 

• 84 

• 84 

.84 
.84 
.84 

.84 
•84 
.84 
.84 
.84 
.84 
.84 

.84 

.84 
.85 
.85 

• 85 

• 85 

• 85 

• 85 

• 85 

• 85 



D. 



9.584177 
584555 

584932 
585309 
585686 
586062 
586439 
5868 1 5 
587190 
587366 
587941 

9 .5383i6 
5SS691 
589066 
589440 
58 9 8i4 
5 9 oi88 
5go562 
590935 
59130S 
5 9 i68 1 

g- 592054 
592426 
692798 
598170 
593542 
593914 
5 9 4285 
5g4656 
595027 
595398 

9.596768 
5 9 6i38 
5g65o8 
596878 

597247 
597616 
597985 
c>98354 
698722 
599091 

9-W59 
599827 
600194 
6oo562 
600929 
60129') 
601662 
602029 
60289) 
602761 

9-6o3i27 
6o3i93 
6o3858 
604223 
6o4588 
604953 
6o5 5 1 7 
6o5682 
606046 
606410 



Sine 



D. 



Cotang. 



6-29 
6-29 
6-28 
6-28 
6-27 
6-27 
6-27 
6-26 
6-26 

6-22 

6-25 

6-25 
6-24 
6-24 
6-23 
6-23 
6-23 

6-22 
6-22 
6-22 
6-21 

6-21 
6. 20 
6- 20 

6- 19 
6-19 
6-i8 
6-i8 
6-i8 
6- 17 
6- 17 

6-17 
6-16 
6- 16 
6-i6 
6-i5 
6-i5 
6-i5 
6-14 
6-i4 
6-i3 

6-i3 
6- 13 

6>I2 
6-12 

6- 11 

6- 1 1 
6-n 
6- 10 
6- 10 
6- 10 

6-09 
6-09 
6-09 
6-o8 
6-o8 
6-07 
6-07 
6-07 
6-06 
6-o6 



10 



10 



Cotang. 

!0-4i5823 
4i5445 
4006S 
414691 
4U3i4 
41393S 
4i356i 
4i3i85 
412810 
412434 
41 2o5g 

41 1684 
41 1309 
410934 
410360 
410186 
409812 
409438 
409065 
408692 
408319 

•407946 
407374 
407202 
406829 
406453 
406086 
4057 1 5 
4o5344 
4o4973 
404602 

io«4o4232 
4o3862 
403492 
4o3i22 
402753 
402384 
4o2oi5 
401646 
401278 
400909 

io-4oo54i 
400173 
399806 
399438 
39907 1 
398704 
3 9 83 38 
397971 
3976.) ) 
397239 

396873 
3i/)5o7 
39614a 
395777 
39541a 
395o47 
3 9 4683 
394318 
l 9 3o54 
393590 



10 



D. 



60 

5n 

58 

57 
56 
55 

54 
53 

32 
31 

1 5o 

4 2 

48 

47 
46 
45 
44 
43 
42 
41 
40 



37 

36 
35 

34 
33 

32 

3i 
3o 

29 
28 
27 
26 

25 

24 

23 
22 
21 
20 

I| 

H 
16 

i5 

H 
i3 
12 
1 1 
10 



7 
6 
5 
4 
3 
2 
1 
o 



Tang. 



M. 



(68 DEGREES.) 



40 



(22 DEGREES.) A TABLE OF LOGARITHMIC 



! m:. 

o 


Sine 


D. 


Cosine 


D. 

.85 


Tang. 


D. 


Cotang. 




9 073575 


5-21 


9-967166 


9-606410 


6-o6 


10-393590 


60 


I 


5 7 3888 


5 


20 


9671 i5 


-85 


606773 


6 


06 


393227 


5o 


2 


574200 


5 


20 


967064 


.85 


607137 


6 


o5 


3 9 2863 


58 


3 


574512 


5 


x 9 


967013 


-85 


607500 


6 


o5 


392500 


57 


4 


574824 


5 


l 9 


966961 


• 85 


607863 


6 


04 


392137 


56 


fi 
& 


575i36 


5 




966910 


-85 


608225 


6 


04 


391775 


55 


6 


575447 


5 


18 


9 6685 9 


-85 


6o8588 


6 


04 


391412 


54 


I 


575 7 58 


5 


18 


966808 


-85 


6o8o5o 


6 


o3 


391050 


53 


576069 


5 


H 


966756 


-86 


609312 


6 


o3 


390688 


52 


9 


576379 


5 


H 


966705 


• 86 


609674 


6 


o3 


390326 


5i 


IO 


576689 


5 


16 


9 66653 


• 86 


6ioo36 


6 


02 


389964 


5o 


CI 


9-576999 
577309 


5 


16 


9 966602 


-86 


9-610397 


6 


02 


10-389603 


4o 


12 


5 


16 


96655o 


•86 


610709 


6 


02 


389241 


48 


i3 


577618 


5 


i5 


966499 


-86 


611120 


6 


01 


38888o 


47 


14 


577927 


5 


i5 


966447 


• 86 


61 1480 


6 


01 


388520 


46 


i5 


5 7 8236 


5 


14 


966395 


• 86 


611841 


6 


01 


388i5 9 


45 


16 


5 7 8545 


5 


14 


966344 


• 86 


612201 


6 


00 


387799 


44 


'7 


5 7 8853 


5 


i3 


966292 


• 86 


6i256i 


6 


00 


38 7 43 9 


43 


18 


579162 


5 


i3 


966240 


• 86 


612921 


6 


00 


387079 


42 


'9 


579470 


5 


i3 


966188 


• 86 


613281 


5 


99 


386719 


41 


20 


5 79777 


5 


12 


966136 


• 86 


6i364i 


5 


99 


38635 9 


40 


21 


9-58oo85 


5 


12 


9-966085 


.87 


9-614000 


5 


98 


10 -386000 


3 9 


22 


58o3g2 


5 


11 


966033 


.87 


614359 


5 


98 


385641 


38 


23 


58o6 9 o 


5 


11 


965981 


.87 


614718 


5 


98 


385282 


37 


24 


58ioo5 


5 


11 


965928 


•87 


615077 


5 


97 


384923 


36 


25 


58i3i2 


5 


10 


965876 


.87 


61 5435 


5 


97 


384565 


35 


26 


58i6i8 


5 


10 


965824 


•87 


615793 


5 


97 


384207 


34 


27 


581924 


5 


09 


965772 


•87 


6i6i5i 


5 


96 


383849 


33 


28 


582229 


5 


09 


965720 


•87 


616509 


5 


96 


383491 


32 


29 


582535 


5 


09 


9 65668 


•87 


616867 


5 


^ 


383i33 


3i 


3o 


582840 


5 


00 


9656i5 


•87 


617224 


5 


9 5 


382776 


3o | 


3i 


9-583U5 


5 


08 


9-965563 


.87 


9-6i 7 582 


5 


9 5 


10-382418 


20 


32 


583449 


5 


07 


9655i 1 


.87 


6i 79 3 9 


5 


9 5 


382061 


28 


33 


583754 


5 


07 


9 65458 


.87 


618295 


5 


9 4 


38i7o5 


27 


34 


584o58 


5 


06 


965406 


.87 


6i8652 


5 


94 


38i348 


26 


35 


58436i 


5 


06 


965353 


• 88 


619008 


5 


94 


380992 


2D 


36 


584665 


5 


06 


9653oi 


-88 


619364 


5 


9 3 


38o636 


24 


3i 


684968 


5 


o5 


963248 


• 88 


61972 1 


5 


9 3 


380279 


23 


38 


580272 


5 


o5 


. 965195 


-88 


620076 


5 


9 3 


379924 
379568 


'J'j 


3 9 


585574 


5 


04 


9 65i43 


-88 


620432 


5 


92 


21 


4o 


585877 


5 


04 


965090 


• 88 


620787 


5 


92 


379213 


20 


4i 


9-586i79 


5 


o3 


9-965037 


• 88 


9-621142 


5 


92 


10-378858 


IQ 


42 


586482 


5 


o3 


964984 


• 88 


621497 


5 


91 


3785o3 


10 


43 


586783 


5 


o3 


96493 1 


• 88 


6218D2 


5 


9 1 


378148 


17 


44 


587085 


5 


02 


964879 


• 88 


622207 


5 


90 


377793 


l6 


45 


587386 


5 


02 


964826 


• 88 


622561 


5 


90 


377439 


i5 


46 


587688 


5 


01 


964773 


• 88 


622915 


5 


90 


377085 


14 


% 


587989 


5 


01 


964719 


• 88 


623269 


5 


89 


376731 


i3 


588289 


5 


01 


964666 


.89 


623623 


5 


89 


376377 


12 


49 


5885 9 o 


5 


00 


964613 


.89 


623976 


5 


89 


376024 


11 


5o 


588890 


5 


00 


964560 


.89 


62433o 


5 


88 


375670 


10 


5i 


9-589190 


4 


99 


9-964507 


.89 


9-624683 


5 


88 


10-375317 


9 


52 


589489 


4 


99 


964454 


.89 


625o36 


5 


88 


374964 





53 


589789 


4 


99 


964400 


.89 


625388 


5 


87 


374612 


7 


54 


590088 


4 


98 


964347 


.89 


625741 


5 


87 


374259 


5 


55 


690387 


4 


98 


964294 


.89 


626093 


5 


87 


073907 


5 


56 


590686 


4 


97 


964240 


.89 


626445 


5 


86 


373D55 


4 


iz 


590984 


4 


97 


964187 


.89 


626797 


5 


86 


373203 


3 


591282 


4 


•97 


964i33 


.89 


627149 


5 


86 


372851 


2 


59 


591580 


4 


96 


964080 


.89 


627501 


5 


85 


372499 
372148 


1 


6c 


591878 


4.96 


964026 


• 8 9 


627852 


5 85 



M. 




Cosine 


D. 


Sine 


D. 


Cotang. 


D. 


Tang. 



(67 DEGREES.) 



&TNES AND TANGENTS. (23 DEGREES.) 



41 



M. 1 


Sine 


D. | 


ol" 


9.591878 


4.96 


I s 


D9.2176 


4.90 


2 


592473 ! 


4.95 


3 


092770 


4.95 


4 


593067 


4-94 


5 


5g3363 


4-94 


6 


593609 


4- 9 3 


7 


593955 


4-9 3 


8 


5g425i 1 


4- 9 3 


9 


594547 | 


4.92 


10 


594842 


4-92 


n 


9.595137 


4-91 


12 


590432 


4-91 


i3 


595727 


4-91 


U 


596021 


4-90 


ID 


5963i5 


4-90 


:6 


596609 


4-89 


3 


596903 


4-8 9 


697196 


4.89 


J 9 


597490 


4-88 


20 


597783 


4-88 


21 


9-598075 


4-8 7 


22 


598368 


4-8 7 


23 


598660 


4-8 7 


24 


598952 


4-86 


25 


599244 


4-86 


26 


. 599536 


4-85 


27 


599827 


4-85 


28 


600118 


4-85 


29 


600409 


4-84 


3o 


600700 


4-84 


3i 


9.600990 


4-84 


32 


6012I0 


4-83 


33 


601570 


4-83 


34 


601860 


4-82 


35 


602 1 5o 


4-82 


36 


602439 


4-82 


ll 


602728 


4-8i 


603017 


4-81 


3q 


6o33o5 


4-8i 


4o 


603594 


4-8o 


4i 


9-6o3882 


4-8o 


42 


604170 


4-79 


43 


604437 


4-79 


44 


604740 


4-79 


45 


6ooo32 


4-78 


46 


6oo3ig 


4-78 


47 


600606 


4-78 


48 


605892 


4-77 


49 


606179 


4-77 


5o 


606465 


4-76 


5i 


9 606701 


4-76 


52 


607036 


4.76 


53 


607322 


4-75 


54 


607607 


4-75 


55 


607892 


4-74 


56 


608177 


4-74 


tl 


608461 


4-74 


608745 


4-73 


5o 


609029 
6093 1 3 


4-73 


6o 


4-73 



Cosine 



9-964026 
963972 
963919 
963865 
9 638u 
963757 
963704 
96365o 
963096 
963042 
963488 

9.963434 
963379 
963320 
963271 
963217 
9 63i63 
963108 
963o54 
962999 
962942 

9-962890 
9 62836 
962781 
962727 
962672 
962617 
962562 
962008 
962453 
962398 

9-962343 
962288 
962233 
962178 
962123 
962067 
962012 
961907 
961902 
96 1 846 

9-961791 

961730 
961680 

961624 
961069 
96 1 5 1 3 
9 6i458 
961402 
96 1 346 
96 1 290 

9-961235 
96 1 1 79 
961 123 
0io67 
96 1 1 I 
96 



96084 
960786 

960730 



D. 

.89 
.89 
.89 
.90 
.90 
.90 

• 90 
.90 
.90 
.90 

• 90 

.90 
.90 
.90 
.90 

• 90 

• 90 
.91 
.91 
.91 
.91 

•9i 
.91 
.91 
.91 

• 91 
.91 
.91 
.91 
.91 
.92 

.92 
.92 
.92 
.92 
.92 

•92 
.92 
.92 
.92 
.92 

.92 
.92 
.92 

■93 
•9 3 
•93 
- 9 3 

•93 
•93 
•93 

- 9 3 
•93 

•93 
•93 
.93 

•93 
•94 

•94 
•94 



Tansr. 



D. 



9-627852 
628203 
628554 
628905 
629255 
629606 
629956 
63o3o6 
63o656 

- 63ioo5 
63i355 

9-631704 
632oo3 
6324oi 
632760 
633o 9 3 

633447 
633795 
634 U3 
634490 
634833 

9>635i85 
635532 
635879 
636226 
636572 
636919 
637260 
63761 1 
637906 
638302 

9-638647 
638992 
639337 
639682 
640027 
640371 
6407 1 6 
641060 
641404 
641747 

9-642091 

642434 

642777 
643i20 

643463 
6438o6 
644 1 i8 
644490 
64483a 
645174 

9-645516 
645857 
646*99 
6465/.0 
646881 
647222 
647562 
647903 
648243 
648583 



5-85 
5-85 
5-85 
5-84 
5-84 
5- S3 
5-83 
5-83 
5-83 
5-82 
5-82 

5-82 
5-8i 
5-8i 



D 

5 
5 
5 
5 
5 
5 

5 
5 

5 

5- 

5- 

5- 

5' 

5- 

5' 

5 

5 
5 
5 
5 
5 
5 
5 
5 
5 
5 



81 

80 
80 

80 

79 
'79 
'79 

.78 
.78 

-78 
•77 
•77 
•77 
•77 
.76 
.76 
•76 

75 
75 
75 
74 
74 
'74 
'73 
■73 
•73 
■72 



5- 

5- 

5- 

5' 

5 

5 

5 



5 

5 



Cotang. 

10-372148 
371797 
371446 
37 1 090 
370745 
370394 
3700.44 
369694 
36o344 
368995 
368645 

10-368296 

367947 
367099 
367200 
366902 
366553 
•3662o5 
365857 
3655io 
365 162 



72 
72 

72 

7i 

71 

71 
70 

70 

70 

69 



1 



10 



IG 



10 



60 

5a 
58 

57 
56 
55 
54 
53 

52 

5i 
5o 

4Q 

48 

47 
46 
45 
44 
43 
42 
41 
40 



3648i5 

364468 

264121 

363 77 4 

363428 

363o8i 

362735 

762389 

362044 

36 1 698 

36i353 
361008 
36o663 
36o3i8 
309973 
309629 
359284 
358940 
358596 
358253 

357909 
357566 
357223 
356880 
356537 
356194 
355852 
3555io 
355i68 
354826 



3 2 
38 

37 
36 
35 

34 
33 

32 

3i 

3o 

29 
28 

27 
26 

25 

24 

S3 
22 
21 

20 

19 
10 

16 
i5 
14 
i3 
12 
11 
10 



Co6ine 



D. 



Sine D. Cotang. 



5-69 


10-354484 


5-69 


304143 


5-6o 
5-6* 


3538oi 


353460 


5-68 


353 1 19 


5-68 


352 


5-67 


353 . 


5-67 


302097 


5-67 


3 017)7 


5-66 


35i4n 



6 
5 
4 
3 
2 
1 
o 



I). 



J Ta ng. I M. ) 



(66 DEGREES.) 



42 



(24 DEGREES.) A TABLE OF LOGARITHMIC 






Sine 


D. 


Cosine 


D. 


Tang. 


D. 


Cotang. 


60 


9 -6093 i 3 


4-73 


9-960730 


•94 


9-648583 


5-66 


10-351417 


i 


609697 


4 


.72 


960674 


.94 


648923 


5 


■ 66 


351077 


5g 


2 


609880 


4 


• 72 


960618 


•94 


649263 


5 


-66 


350737 


58 


3 


610164 


4 


• 72 


960561 


.94 


649602 


5 


■ 66 


35o398 


12 


4 


610447 


4 


•7i 


96o5o5 


•94 


649942 


5 


•65 


35oo58 


5 


610729 


4 


•7i 


960448 


•94 


65o28i 


5 


•65 


349719 


55 


6 


61 1012 


4 


•70 


960392 


•94 


65o62o 


5 


•65 


34938o 


54 


7 


61 1 294 


4 


■ 70 


96o335 


.94 


650959 


5 


•64- 


349041 


53 


c 


61 1 D76 


4 


• 70 


960279 


•94 


65 1 297 


5 


■64 


34o7u3 


5a 


9 


6n858 


4 


.69 


960222 


•94 


65 1 636 


5 


■64 


348364 


5i 


10 


612140 


4 


.69 


960165 


.94 


651974 


5 


•63 


348026 


5n 


ii 


9-612421 


4 


■ 69 


9-960109 


- 9 5 


9-6523i2 


5 


■63 


10-347688 


49 


12 


612702 


4 


-68 


960052 


• 9 5 


65265o 


5 


■63 


34735o 


48 


t3 


612983 


4 


■ 68 


959995 


• 9 5 


652988 


5 


•63 


347012 


47 


U 


.613264 


4 


.67 


959988 


• 9 5 


653326 


5 


•62 


346674 


46 


i5 


6i3545 


4 


.67 


959882 


• 9 5 


653663 


5 


•62 


346337 


45 


16 


6i3825 


4 


.67 


959825 


• 9 5 


654ooo 


* 
u 


•62 


346000 


44 


17 


6i4io5 


4 


-66 


959768 


• 9 5 


654337 


5 


•61 


345663 


43 


18 


6i4385 


4 


• 66 


9597 1 1 


• 9 5 


654674 


5 


•61 


345326 


42 


19 


6i4665 


4 


66 


959654 


• 9 5 


655on 


5 


•61 


344989 


4i 


20 


614944 


4 


65 


959596 


- 9 5 


655348 


5 


•61 


344652 


40 


21 


9-6i5223 


4 


65 


9-959539 


. 9 5 


9-655684 


5 


•60 


io-3443i6 


It 


22 


6i55o2 


4 


65 


959482 


• 9 5 


656o2o 


5 


■60 


343980 


23 


6i5 7 3i 


4 


64 


959425 


- 9 5 


656356 


5 


■60 


343644 


u 


24 


616060 


4 


64 


9 5 9 368 


. 9 5 


606692 


5 


-5 9 


3433o8 


23 


6;5338 


4 


64 


959310 


.96 


657028 


5 


-5 9 


342972 


3i 


26 


616616 


4 


63 


959253 


.96. 


657364 


5 


-5 9 


342636 


34 


27 


616894 


4 


63 


959195 


.96 


657699 


5 


■5o 


3423oi 


33 


28 


617172 


4 


62 


959138 


• 96 


658o34 


5 


■58 


341966 


33 


3 9 


617450 


4 


62 


959081 


.96 


65836 9 


5 


• 58 


34i63i 


3i 


3o 


617727 


4 


62 


959023 


.96 


658704 


5 


•58 


341296 


3o 


(I 


9-618004 


4 


61 


9 058965 


.96 


9-659039 


5 


■ 58 


10-340961 


28 


J2 


618281 


4 


61 


958008 
95Sfyio 


.96 


659373 


5 


■57 


340627 


20 


33 


6i8558 


4 


61 


.96 


659708 


5 


•5 7 


340292 


2 


U 


6i8834 


4 


60 


958792 


.96 


660042 


5 


57 


339953 


J5 


619110 


4 


60 


958734 


.96 


660376 


5 


■ 5 7 


339654 


25 


36 


6i 9 386 


4 


60 


958677 


.96 


660710 


5 


56 


339290 

338907 


24 


U 


619662 


4 


5 9 


958619 


.96 


661043 


5 


■ 56 


23 


6i 99 38 


4 


5 9 


95856i 


.96 


661377 


5 


■ 56 


338623 


2? 


3 9 


620213 


4 


5 9 


9585o3 


•97 


661710 


5 


55 


338290 


21 


4o 


620488 


4 


58 


958445 


•97 


662043 


5 


55 


337907 


2D 


4i 


9-620763 


4 


58 


9-958387 


•97 


9-662376 


5 


55 


10*337624 


IO 


42 


62io38 


4 


57 


958329 


•97 


662709 


5 


54 


337291 


IO 


43 


62i3i3 


4- 


5 7 


958271 


•97 


663o42 


5 


54 


336g58 


17 


44 


62i58 7 


4- 


57 


9582i3 


•97 


6633 7 5 


5 


54 


336625 


l6 


4 'J 


621861 


4- 


56 


958i54 


•97 


663707 


5 


54 


336293 


i5 


46 


622i35 


4- 


56 


958096 


•97 


664039 


5 


53 


335961 


14 


47 


622409 


4- 


56 


958o38 


•97 


664371 


5 


53 


335629 


i3 


48 


622682 


4- 


55 


957979 


•97 


664703 


5 


53 


335297 


12 


49 


622906 


4- 


55 


957921 


•97 


665o35 


5 


53 


334965 


11 


5o 


623229 


4- 


55 


957863 


•97 


665366 


5 


52 


334634 


10 


5t 


9-6535o2 


4- 


54 


9-957804 


•97 


9*665697 


5 


52 


io-3343o3 


I 


02 


623774 


4- 


54 


957746 


.98 


666029 


5 


52 


333971 


63 


624047 


4- 


54 


957687 


.98 


66636o 


5 


5i 


333640 


7 


54 


624319 


4- 


53 


957628 


.98 


666691 


5 


5i 


3333o9 


6 


55 


624591 


4- 


53 


957570 


.98 


667021 


5- 


5i 


332979 


5 


56 


624863 


4- 


53 


95751 1 


.98 


667352 


5- 


5i 


332648 


4 


57 


625i35 


4- 


52 


957452 


.98 


667682 


5- 


5o 


3323i8 


3 


58 


625406 


4« 


52 


957393 


.98 


6680 1 3 


5- 


5o 


331987 


2 


59 


625677 
6a5 9 48 


4- 


52 


9 5 7 335 


.98 


668343 


5- 


5o 


33i657 


I 


6o 


4-5i 


957276 


.98 


668672 


5-5o 


33i328 




M. 1 


Cosine 


D. 


Sine 


D. 


Cotansf. 


D. 1 


Tang. 



(65 DEGREES.) 



SINES AND TANGENTS. (25 DEGREES.) 



48 



[""ML. 



2 

3 
4 
5 
6 



Sine 



D. 



9 
lb 

II 
IS 

i3 

U 
i5 

16 

H 
18 

'9 

20 

21 
2? 
23 
24 
25 
26 

27 
23 
29 
3o 

3i 

'SI 

33 

34 
35 
36 

37 
38 

3 9 

4o 

4i 
42 
43 
44 
45 
46 

47 
48 

49 
5o 

Si 

52 

53 
54 
55 



9-6259 i3 
626219 
626490 
626760 
627030 
627300 
627370 
627840 
628109 
628378 
628647 

a S28916 
629185 
6?9452 
629P1 
629989 
630257 
63o524 
630792 
63 10 59 
63i326 

^•63 1 593 
63r85cj 

63'2I2D 

632392 
632658 
632923 
633i8 9 
633454 
633719 
6339S4 

9-6342.49 
6345 i 4 
634778 
635o42 
6353 06 
635570 
635834 
636097 
63636o 
636623 



4-5i 
4-5i 
4-5i 
4-5o 
4-5o 



9-6368% 
637148 

63741 1 

637673 

637935 

638197 

638458 

63*720 

638 9 8i 

639242 

9-6395o3 
639764 
640024 
640284 
640644 



56 


640804 


57 
58 


64 1 064 


641324 


5g 


641 584 


60 


641842 


Cosine 



Cosine 



D. 



Tan?- 



A- 
4' 
4' 
4- 
4- 
4- 

4- 
4- 
4- 
4- 
4- 
4' 
4' 
4 
4 
4 



oo 

49 
49 
49 

48 

48 

47 
47 
'47 
.46 
■ 46 
.46 
.46 
•45 
•45 
45 



4-44 
4-44 
4.44 
4.43 



43 
43 
42 
42 
42 
41 



1 



41 

40 
40 
40 
3 9 
3 9 
3 9 
4-38 
4-38 
4-33 

4-3 7 
4-37 
4-37 
4-37 
4-36 
4-36 
4-36 
4-35 
4-35 
4-35 

4-34 
4-34 
4-34 
4.33 
i-33 
4-33 
4-32 
i-32 

1-32 

4 3i 



? - 957276 
957217 
937158 
957099 
957040 
9 56 9 8i 
956921 
956862 
9 568o3 
956744 
9 56684 

9.956625 
g56566 
9565o6 
956447 
95638 7 
906327 
9 56268 

g56208 

956148 
956089 

9.956029 
955969 
955909 
955849 

955789 

955729 
955669 
955609 

955548 
955488 

1-955428 

9 55368 
955307 

955247 
9 55i86 
955126 
955o65 
955oo5 

954944 

954883 

9-954823 
954762 
904701 
954640 
95457Q 
954")i8 

954457 

954396 
954335 
954274 

9-954213 
954i52 
954090 
954029 
953968 
953fjo6 
953845 
953783 
953722 
953660 



.98 
.98 

"% 

-o3 

•98 

•99 
•99 
•99 
•99 
•99 

•99 
•99 
•99 
•99 
•99 
•99 

•99 

1 -oo 

I -00 
I -00 



00 
00 

00 



I -00 
I -00 
I -00 



•00 
•00 



D. 



Cotan£. 



1 -oo 

I -00 

1-01 
I -01 
I -01 
I-OI 
I -01 
I -01 
I-OI 
I-OI 
I-OI 
I-OI 

I-OI 

I -OI 
I -01 
I -01 
I -01 
I -02 
I -02 
I -02 
I -02 
I -02 

I -02 



•02 
•02 



I -02 



I -02 

I -02 
1 -02 

1 -02 
I-t.3 

i-(3 



9.668673 
669002 
669332 
669661 
669991 
670320 
670649 
670977 
671306 
671634 
671963 

9-672291 
672619 

672947 
673274 
673602 
673929 
674257 
674584 
674910 
675237 

9-675564 
675890 
676216 
676543 
676869 

677194 
677520 
677846 
67817 1 
678496 

9-678821 
679146 

679471 
679795 
680120 

68o444 
680768 
681092 
681416 
681740 

9-682063 
682387 
682710 
683o33 
683356 
683679 
684001 

684324 

684646 
684968 

9-685290 
685612 
685 9 3i 
686255 
6865 77 
686898 
687219 
68 7 5 40 
687861 
683 \ 82 



5-5o 
5-49 
5-49 
5-49 
5-43 
5-48 
5-48 
5-48 
5-47 
5-47 
5-47 

5-47 
5-46 
5-46 
5-46 
5-46 
5-45 
5-45 
5-45 
5-44 
5-44 

5-44 
5-44 
5-43 
5-43 



D. 



Sine 



D. 



io-33i327 I 60 

330998 1 5o 

33o668 I 58 

33o339 57 

330009 56 

329680 55 

32 9 35i 54 

329023 53 

328694 52 

328366 5i 

328037 5o 

10-327709 4Q 

327381 48 

327053 47 

326726 46 

3263 9 8 45 

326071 44 

325 7 43 43 

325416 42 

325090 41 

324763 40 

10-324436 39 

324110 38 

323784 37 

323457 36 

5-43 3?3i3i 35 

5.43 322806 34 

5-42 322480 33 

5.42 322i54 32 

5-42 321829 3i 

5.42 32i5o4 3o 

5-4i io-32ii79 2 9 
5.41 320854 28 
5.41 320529 27 

5.41 320205 26 

5-4o 319880 25 
5-4o 3i9556 24 

5.40 310232 23 

5-40 318908 22 
5-39 3 1 8584 21 
5-3g 318260 20 

5-39 10-317937 19 

5-39 3h6i3 18 

5-38 317290 17 

5-38 316967 16 

5-38 3 1 66 14 i5 

5-38 3i632i 1 4 

5-37 3i5 9 99 l3 

5-37 3i5676 12 

5-37 3 l 5354 1 M 

5.37 3i5o32 

5-36 io-3i47'o 

5-36 3i4388 

5-36 3i4o66 

5-36 3 1 374 5 

5-35 3 1 3423 

5-35 3i3io2 

5-35 312781 

5-35 312460 

5-34 3i2-3o 

5-34 3n»i8 



Uotanjr. 



I). 



Tane. 



I 

1 
6 

D 

4 
3 
2 
1 

o 



(61 DEGREES.) 



(26 DEGREES.) A TABLE OF LOGARITHMIC 



M. 




Sine 


D. 


Cosine I 


). 


Tang. 


D. 


Cotang. 


60 


9-641842 


4-3i 


9-953660 1 


o3 


9-688182 


5-34 


io-3ii8i8 


i 


642101 


4 


■3i 


953599 1 


o3 


688502 


5 


34 


311498 


5 9 


2 


642360 


4 


•3i 


953537 1 


o3 


688823 


5 


34 


311177 


58 


3 


642618 


4 


3o 


953475 1 


o3 


689143 


5 


33 


310837 


57 


4 


642877 


4 


3o 


9534i3 1 


o3 


68 9 463 


5 


33 


3io537 


56 


5 


643i35 


4 


-3o 


953352 1 


o3 


689783 


5 


33 


310217 


55 


6 


643393 


4 


3o 


953290 1 


o3 


690103 


5 


33 


309897 


54 


7 


64365c 


4 


29 


953228 1 


o3 


690423 


5 


33 


309577 


53 


8 


643908 


4 


29 


953166 1 


o3 


690742 


5 


32 


309258 


52 


9 


644i65 


4 


29 


953io4 1 


o3 


691062 


5 


32 


3oS 9 38 


5i 


10 


644423 


4 


-28 


953042 1 


o3 


691381 


5 


32 


308619 


5o 


ii 


9-644680 


4 


28 


9-952980 1 


-04 


9-691700 


5 


3i 


io-3o83oo 


4o 


12 


644936 


4 


28 


952918 1 
952855 1 


04 


692019 


5 


3i 


307981 


48 


i3 


645193 


4 


27 


04 


692338 


5 


3i 


307662 


47 


U 


645430 


4 


27 


952793 1 


04 


692656 


5 


3i 


307344 


46 


i5 


645706 


4 


27 


952701 1 


04 


692975 


5 


3i 


307025 


45 


16 


645962 


4 


26 


952669 1 


04 


693293 


5 


3o 


306707 


44 


17 


646218 


4 


26 


952606 1 


04 


693612 


5 


3o 


3o6388 


43 


18 


646474 


4 


26 


952544 1 


04 


693930 


5 


3o 


306070 


42 


x 9 


646729 


4 


25 


952481 1 


04 


694248 


5 


3o 


3o5752 


4i 


20 


646984 


4 


25 


952419 1 


04 


694566 


5 


29 


3o5434 


4o 


21 


9-647240 


4 


25 


9-952356 1 


04 


9 -6 9 4883 


5 


29 


ic-3o5ii7 


3 9 


22 


647494 


4 


24 


952294 1 


04 


695201 


5 


29 


304799 


38 


23 


°47749 


4 


24 


9D2231 1 


04 


695518 


5 


29 


3o4482 


37 


24 


648004 


4 


24 


952168 1 


•o5 


6 9 5836 


5 


29 


304164 


36 


25 


648258 


4 


24 


952106 1 


o5 


696153 


5 


28 


3o3847 


35 


26 


6485 1 2 


4 


23 


952043 1 


o5 


696470 


5 


-28 


3o353o 


34 


27 


648766 


4- 


23 


951980 1 


o5 


696787 


5 


•28 


3o32i3 


33 


28 


649020 


4 


23 


951917 1 


o5 


697103 


5 


28 


302807 


32 


29 


649274 


4 


22 


95i854 1 


o5 


697420 


5 


27 


3o258o 


3i 


3o 


649527 


4 


22 


951791 1 


o5 


697736 


5 


27 


3o2264 


3o 


3i 


9-649781 


4 


22 


9-951728 1 


o5 


9-698053 


5 


27 


10.301947 


29 


32 


65oo34 


4 


22 


95i665 1 


o5 


698I69 


5 


27 


3oi63i 


28 


33 


650287 


4 


21 


951602 1 


o5 


698683 


5 


26 


3oi3i5 


27 


34 


65o539 


4 


21 


95i539 1 


o5 


699001 


5 


26 


300999 


26 


35 


650792 


4 


21 


951476 .1 


o5 


6993 1 6 


5 


26 


300084 


25 


36 


65 1 044 


4 


20 


951412 1 


o5 


69963 2 


5 


26 


3oo368 


24 


37 


65i 297 


4 


20 


901349 1 


06 


699947 


5 


26 


3ooo53 


23 


38 


65i 549 


4 


20 


951286 1 


06 


700263 


5 


25 


299737 


22 


3 9 


65i8oo 


4 


10 


951222 1 


06 


700578 


5 


25 


299422 


21 


4o 


652052 


4 


J 9 


95 1 1 59 1 


06 


7CJ893 


5 


25 


299107 


20 


4i 


9>6523o4 


4 


IQ 


9-951096 1 


06 


9-701208 


5 


24 


10.298792 


Io 


42 


652555 


4 


18 


95io32 1 


06 


70i523 


5 


24 


298477 


10 


43 


6528o6 


4 


18 


950968 1 


06 


701837 


5 


24 


? 9 8i63 


17 


44 


653o57 


4 


18 


950905 1 


06 


702152 


5 


24 


, 297848 


16 


45 


6533o8 


4 


18 


•950841 1 


06 


702466 


5 


24 


297534 


i5 


46 


653558 


4 


17 


950778 1 


06 


702780 


5 


23 


297220 


14 


47 


6538o8 


4 


17 


950714 1 


06 


703095 


5 


23 


2Q6905 


i3 


48 


654059 


4 


'7 


95o65o 1 


06 


703409 


5 


23 


290091 


12 


i 9 


654309 


4 


16 


95o586 1 


06 


703723 


5 


23 


296277 


11 


5o 


654558 


4 


16 


95o522 1 


07 


704036 


5 


22 


295964 


10 


5i 


9.654808 


4 


16 


9-95o458 1 


07 


9-704350 


5 


22 


io-29565o 


Q 


52 


655o58 


4 


16 


950394 1 


°7 


704663 


5 


22 


295337 


8 


53 


655307 


4 


i5 


95o33o 1 


07 


704977 


5 


22 


295023 


7 


54 


655556 


4 


i5 


950266 1 


°7 


705290 


5 


22 


294710 


6 


55 


6558o5 


4 


i5 


950202 1 


°7 


7o56o3 


5 


21 


294397 


5 


56 


656o54 


4 


14 


95oi38 1 


07 


705916 


5 


21 


294084 


4 


n 


6563o2 


4 


14 


950074 1 


07 


706:; 2 8 


5 


21 


293772 


3 


65655i 


4 


14 


95ooio 1 


07 


706541 


5 


21 


293459 


2 


59 


656799 


4 


i3 


9499 / .5 1 


07 


706854 


5 


21 


293146 


1 


60 


657047 


4-i3 


949881 1 


07 


707166 


5-20 


292834 




M. 


Cosine 


D. 


Sine I 


). 


Cotansy. 


D. 


Tang. 



(G3 DEGREES.) 



SINES AND TANGENTS. (27 DEGREES.) 



45 



M. 

o 
i 

2 

3 
4 
5 
6 



Sine 



D. 



9 
ic 

ii 

12 

i3 
U 
i5 
16 

18 

19 
20 

21 
22 

23 

24 

23 

26 

27 
28 

3o 
3i 

32 

33 
34 
35 
36 

37 
38 
3 9 
4o 

41 
42 
43 

44 
45 
46 

47 
48 

5o 
5i 

52 

53 
5.4 
55 
56 



J. 657047 
657295 
657342 
657790 
658o37 
638284 
65853 1 
658778 
659023 
659271 
6595i- 

9-659763 
660009 
660233 
66o5oi 
660746 
660991 
66 1 236 
661481 
661726 
661970 

9.662214 
662459 
662703 
662946 
663 190 
663433 
663677 
663Q20 
664i63 
664406 

9.664648 
664891 
665 1 33 
665375 
6656i7 
665859 
666100 
666342 
666583 
666824 

g. 667 06 5 
6673o5 
667546 
667786 
668027 
668267 

6685o6 
668746 
66S986 
669225 

9 -669 46 i 
669703 
669942 
670181 
670419 
670638 
670896 
671134 
671372 
671609 



4-i3 

4-i3 

4- 

4- 

4' 

4' 

4 



12 

12 
12 
12 
11 



4-u 
4-u 
4- 
4- 



10 
10 



4- 10 
4-09 



09 
09 
09 
08 
08 
08 
4-07 
4-07 

4-07 
4-07 
4-o6 



4- 
4- 
4- 

4' 
4 
4 

4 



Cosine 



D 



Tan< 



D. 



Cotang. 



06 
06 
o5 
o5 
o5 
o5 
04 



4-o4 
4-o4 



©3 
o3 
o3 
02 
02 
02 

4*02 

4-oi 



01 
01 
01 
ot 
00 
00 

99 
99 

'99 
3-99 

3- 9 8 
3.98 
3-98 
3-97 
3-97 

3 97 
3- 97 

3- 9 6 

3- 9 6 

3.96 



9.949881 
949816 
949732 
949688 
949623 
949558 
949494 
949429 
949364 
949300 
949235 

9-94917? 
949103 

949040 
948975 
948910 
948845 
948780 
948713 
94865o 

948584 

9-948519 
948454 
948388 
9 48323 
948257 
948192 
948126 
948060 

9 4799 5 
947929 

7.947863 

947797 
947731 
947663 
947600 
947533 

947467 
947401 
947333 
947269 

9.947203 

947" 6 

947070 

947004 
946937 
94687 1 
946804 
946738 
94667 1 
946604 

9-946538 
9 4647 1 
9 46404 

946337 
946270 
946203 
9'/) 1 36 
946069 
946002 
943935 



1-07 
1-07 



07 
08 
oS 
08 
08 
08 
08 
08 
■o8 



1-08 
1-08 
1-08 
i-oS 
1-08 
i- 08 
1 -09 
1-09 
1-09 
1-09 

1-09 
1-09 
1 -09 
1-09 
1 -09 
1-09 
1 -09 
1 -09 

1 -10 
I-IO 



I 
I 

I • 
I • 
I • 
I • 
I • 
I • 
I • 
I • 

I • 
I ■ 
I • 

1 ■ 
I 

I 
I 
I 

I 
I 



10 

1 1 
1 1 
1 1 
1 1 
1 1 
1 1 
1 1 
1 1 
1 1 

1 1 

• II 
■ 1 1 

• 1 1 

• 12 

• 12 

• 12 

•12, 
12 
12 



I • 


10 


I ■ 


IC 


I 


10 


I 


10 


I 


10 


I 


10 


I 


• 10 


I 


• 10 


I 


• 10 


I 


• 10 



9.707166 

707478 
707790 

708102 
708414 
708726 
709037 
709349 

709660 

709971 

710282 

J. 710593 
710904 
7 1 1 2 id 

71 ID25 

7 u836 
7 1 2 1 46 
7 1 2456 
712766 
713076 
7 i3386 

9. 7 13696 
7i4oo5 
7i43i4 
714624 
7 1 4933 
715242 
7 1 555 1 
7i5S6o 
716168 

716477 
9.716785 
717093 

717401 
717709 
718017 
7i8325 
7 i8633 
718940 
719248 
719555 

9.719862 
720169 
720476 
720783 
721089 
721396 
721702 
722009 
722113 
722621 

9.722927 
723232 
723538 
723844 

7241/9 
724454 
72475Q 
72 )()')> 
725369 
72)674 



5-20 
5-20 
5-20 
5-20 

5-19 
5-19 
5-19 
5-19 
5-19 
5-i8 
5-i8 



10-292834 
292522 
292210 
291898 
291586 
291274 
290963 
290631 
290340 
290029 
289718 



5-i8 
5-i8 
5- 18 
5.17 
5-H 
5-17 
5- 17 
5-i6 
5.16 
5-i6 

5-i6 
5-i6 
5-i5 
5-i5 
5-i5 
5-i5 
5i4 
5-i4 
5-i4 
5-i4 

5-U 
5-i3 
5-i3 
5-i3 
5-i3 
5-i3 

5-12 

5-12 
5-12 
5- 12 

5-12 

5-u 
5- 11 

5-u 
5-u 
5-u 
5- 10 
5 10 
5- 10 
5-io 

5- 10 
5-09 
5-09 
5-09 
3-09 
5-09 
5-o8 
5-o8 
5-o8 
5-o8 



10 



289407 

289096 

288785 

288475 

288164 

287854 

287544 

287234 

286924 

286614 



io-2863o4 
285993 
285686 
285376 
285067 
284758 

284449 
284U0 

283832 

283323 

io-2832i5 

282907 
282399 
282291 
281983 
281670 
281367 
281060 
280752 
28o445 

io-28oi38 
279831 

279324 
279217 
27891 1 

278604 
278298 
277991 
277685 

277379 

10-277073 
276768 
276462 
276156 
27585i 
275546 
275241 
274g35 

274631 
214326 



60 

59 
58 

57 
56 
55 
54 
53 

52 

5i 

30 

49 
48 

47 
46 
45 

44 
43 
42 
4i 

40 

3 9 
38 

37 
36 
35 

34 
33 

32 

3i 
3o 

2Q 
28 

27 
26 
25 

24 
23 
22 
21 
20 



Cosine 



D. 



Sine 



1). 



Cotang. 



D. 



19 
18 

17 
16 

13 

14 
i3 
12 
u 
10 



; 

6 
5 

4 
3 
2 
1 
o 



Tung. M- _ 



(62 DEGREES.) 



46 



(23 DEGREES.) A TABLE OF LOGARITHMIC 



M. 




Sine 


D. 


Cosine 


D. 


Tang. 


D. 


\ Cotang. 


60 


9-671609 


3- 9 6 


9'945o35 


I«12 


9-720674 


5-o8 


10-274326 


i 


671847 


3-95 


940868 


1-12 


725979 


5- 08 


274021 


5 9 


2 


672084 


3- 9 5 


9458oo 


I-I2 


726284 


5-07 


273716 


58 


3 


672321 


3-95 


945733 


I-I2 


726588 


5-07 


273412 


5 7 


4 


672008 


3-95 


940666 


I-I2 


726892 


5 -07 


278108 


. 56 


5 


672795 


3-94 


945598 


I-I2 


727197 


5-07 


272803 


! 5£> 


6 


673o32 


3- 9 4 


94553 1 


I -12 


727501 


0-07 


272499 


54 


I 


673268 


3-94 


940464 


I-I3 


727805 


5-o6 


272195 


53 


6735o5 


3- 9 4 


945396 


i-i3 


728109 


5- 06 


27 1 89 1 


52 


9 


673741 


3- 9 3 


940328 


i-i3 


728412 


5-o6 


27088 


5i 


10 


673977 


3- 9 3 


940261 


1 -i3 


728716 


5-o6 


271284 


5o 


ii 


9-674213 


' 3- 9 3 


9-940193 


i-i3 


9-729020 


5-o6 


30-270980 


4q 


12 


674448 


3-92 


940125 


i-i3 


729323 


5-o5 


270677 


48 


i3 


674684 


3-92 


940008 


i-i3 


729626 


5-o5 


270374 


47 


14 


674919 


3-92 


944990 


i-i3 


729929 


5-o5 


270071 


46 


i5 


673135 


3-92 


944922 


i-i3 


73o233 


5-o5 


269767 


45 


16 


675390 


3 91 


944854 


i-i3 


73o535 


5-o5 


269465 


44 


1-7 


675624 


3-91 


944786 


i-i3 


73o838 


5-o4 


269162 


43 


18 


675809 


3-91 


944718 


i-i3 


73,ii4i 


5 04 


268809 


42 


'9 


676094 


3-91 


94465o 


i-i3 


731444 


5-o4 


268556 


4i 


20 


676328 


3-90 


944082 


i-i4 


731746 


5- 04 


, 268254 


40 


21 


9-676562 


3-90 


9-9445i4 


1-14 


9' 732048 


5-o4 


10-267952 


39 


22 


676796 


3-90 


944446 


1-14 


73235i 


5-o3 


267649 


38 


23 


677080 


3- 9 o 


944377 


i-i4 


732653 


5-o3 


267347 


37 


24 


677264 


3.69 


944309 


1-14 


782955 


5-o3 


267045 


36 


25 


677498 


3-8 9 


944241 


i-U 


733257 


5-o3 


266743 


35 


26 


677731 


3-8 9 


944172 


1-14 


733558 


5-o3 


266442 


34 


27 


677964 


3-88 


944104 


1-14 


73386o 


5-02 


266140 


33 


28 


678197 


3-88 


944o36 


1-14 


734162 


5-02 


265838 


32 


29 


678430 


3-88 


943967 


1-14 


734463 


5-02 


265537 


3i 


3o 


678663 


3-88 


943899 


I-J4 


734764 


5-02 


265236 


3o 


3i 


9-678895 


3-8 7 


9- 94383o 


I- 14 


9- 735o66 


5-02 


10-264934 


29 


32 


679128 


3-87 


943761 


i-i4 


735367 


5-02 


264633 


28 


33 


. 679360 


3-87 


943693 


i-i5 


735668 


5-oi 


264332 


27 


34 


679592 


3-8 7 


943624 


1 -i5 


735969 


5-oi 


26403 1 


26 


35 


679824 


3-86 


943555 


i-i5 


736269 


5-oi 


263731 


25 


36 


68oo56 


3-86 


943486 


i-i5 


736070 


5-oi 


26343o 


24 


3 7 


680288 


3-86 


943417 


i-i5 


736S71 


5-oi 


263129 


23 


38 


680519 


3-85 


943348 


i-i5 


7371 7 1 


5-oo 


262829 


22 


3 9 


680750 


3-85 


943279 


i-i5 


737471 


5- 00 


262529 


21 


40 


680982 


3-85 


9432io 


i-i5 


737771 


5-oo 


262229 


20 


4i 


9-68i2i3 


3-85 


9-943141 


i-i5 


9-738071 


5-oo 


10-261929 


19 


42 


68i443 


3-84 


943072 


i-i5 


738371 


5-oo 


261629 


10 


43 


681674 


3-84 


943oo3 


i-i5 


738671 


4-99 


261329 


»7 


44 


681905 


3-84 


942934 


i-i5 


738971 


4-99 


261029 


16 


45 


682i35 


3-84 


942864 


ii5 


739271 


4-99 


260729 


i5 


46 1 


682365 


3-83 


942790 


1-16 


739070 


4-99 


260430 


14 


4 7 


682095 


3-83 


942726 


1-16 


739S70 


4.99 


260180 


i3 


48 


682Sz5 


3-83 


942606 


] -16 


740169 


4.99 


259831 


13 


49 


683o55 


3-83 


94258 7 


1 • 16 


740468 


4-98 


259032 


11 


5o 


683284 | 


3-82 


9425i7 


1-16 


740767 


4-98 


259233 


IO 


5 


g-6835l4 1 


3-82 


9-942448 


1-16 


9-741066 


4-98 


10-258934 


Q 


5a 


683743 


3-82 


942378 


1-16 


74 1 365 


4-98 


258635 


8 


53 


683972 


3-82 


9423o8 


1-16 


741664 


4-98 


258336 


7 


54 


684201 


3-8i 


942239 


1-16 


741962 


4-97 


258o38 


6 


55 


68443o 


3-8l 


942169 


1*16 


742261 


4-97 


257739 


5 


56 


684608 


3-8i 


942099 


1 • 16 


742059 


4-97 


257441 


4 


5 7 


684887 


3-8o 


942029 


1 -16 


74285S 


4-97 


257142 


3 


58 


685n5 


3-8o 


94i9 5 9 


1-16 


743 1 56 


4-97 


256844 


2 


5 9 


685343 


3- 80 


941889 


1-17 


743454 


4-97 


256546 


1 


60 


685571 

1 


3-8o 

1 


941819 


1-17 


743752 


4-96 


256248 



M. 




Cosine 


D. I 


Sine 


D. 


Cotang. 


D. 1 


Tang. 



(Gl DEGREES.) 



SINES AND TANGENTS. (29 DEGREES.) 



47 



M. 



Sine 



D. 



Cosine 



o 
i 

2 

3 

4 

5 
6 

1 
8 

9 

10 

ii 

12 

i3 

14 

i5 

16 

n 

18 
19 

20 

21 
22 
23 

24 
23 
26 

2 
2 

29 

30 

3i 

32 

33 
34 
35 
36 

37 
38 

39 
4o 

4l 

42 
43 
44 
45 
46 

a 
u 

52 

53 
54 
55 
56 

57 
58 

6o 



y 68557 i 
680799 
686027 
686254 
686482 
686709 
686 9 36 
687163 
687389 
687616 
687843 

9.68S069 
688293 
688521 

688747 
688972 
689198 
689423 
689648 
689873 
690098 

■690323 
690048 
690772 
690996 
691220 
691444 
691668 
691892 
6921 1 5 
692339 

•692062 
692780 
693008 
693231 
693453 
693676 
693898 
694120 
694342 
694564 

1-694786 
695007 
69 5229 
695450 
690671 
695892 
696 1 1 3 
6 9 o334 
696554 
696775 

) -696995 
697215 
697430 
697604 
607874 
69-1094 
6983i3 
6o8532 
698751 
698970 



3-8o 

3-79 
3-79 

3-79 
3-79 
3-78 
3.78 
3-78 
3-78 

3-77 
3-77 



77 
77 
76 
76 
76 
76 

75 
3-75 
3-75 
3-75 



3-74 
3-74 
3-74 
3-74 
3-73 
3-73 
3-73 
3-73 
3-72 
3.72 

3-72 
3-7i 
•7i 
■7i 
•7i 
•70 
.70 
•70 
3-70 
3-6 9 

3-6 9 

3-69 
•5 /.-. 



68 
68 

68 
• 68 

■67 
67 



3-6 7 

3-67 
3-66 
3-66 
-66 



Cosine 



66 
65 
65 
65 
3-65 
3.64 



D. 



Tang. 



D. 



Cotang. 



D. 



9-941819 

941749 
941679 
941609 
941539 
941469 
941398 
941328 
941208 
94i 187 
941117 

9-941046 
940970 
940900 
940834 
940763 
940693 
940622 
94o5o 1 
940480 
940409 

9'94o33S 
940267 
940 1 96 
940125 
940054 
939982 
93991 1 
939840 
9 3 9 768 
939697 

9-939625 
939004 I 
939482 

939410 I 
939339 i 
939267 i 

93919 5 

939123 I 

939052 j 
938980 

9.93890S 

938836 ; 

o3oT';3 
938f'<9? 
93S619 
938547 

938473 
938402 

93833o 
9 38258 

9 -938 1 85 
$38113 
938040 

937967 

9' ! ~ 
937822 

937749 
937676 
937 

937531 

Sine 



1-17 

1-17 
i- 17 

1-17 
1-17 
1V17 

1-17 
1-17 
1-17 
1-17 
1-17 

i-iS 
i- 18 



18 
18 
18 
18 
18 
1-18 
1-18 
1-18 



1 ■ 


18 


I • 


18 


I • 


18 


I • 


19 


1 


19 


I 


•9 


I 


'9 


I 


'9 


I 


>9 


I 


•9 


I 


19 


I 


•19 


I 


•19 


I 


•19 


I 


•19 


I 


•20 


I 


•20 


I 


•20 


I 


•20 


I 


■20 



I -20 
I -20 
I -20 
I -20 
1-20 
I -20 
I -20 
I 21 
I -21 
I - 2 I 

I -21 
I - 2 I 
1 -21 
I -21 
I -21 
I-2I 
I -21 
I -21 
[•21 
I -21 

D. 



9-743752 
744o5o 
744348 
744645 
744943 
745240 
745533 
745835 
746132 
746429 
746726 

9-747023 

747319 
747616 

7479 l3 
748209 
7485o5 
748801 
749097 
749393 
749689 

9.749985 
750281 
750576 
700872 
761167 
751462 
751707 
752o52 

752347 
752642 

9.752937 
75323i 
703026 
703820 

754i 1 5 
704409 
70470J 

754997 
705291 
755585 

•755878 
706172 
706465 
756759 
707052 
757345 
7 o 7 638 
757931 
738224 
758017 

i-7588io 
709102 
709395 
759687 
759979 
760272 
76006 \ 
760806 
761 1 18 
761439 



4.96 
4.96 
4.96 
4.96 
4.96 
4.96 
4-90 
4- 9 5 
4- 9 5 
4- 9 5 
4- 9 5 



jo 



4 

A- 

4- 

4' 

4' 

4' 

4 

4 

4 

4 

4 

4 

4 

4 

4 

4 

4 

4 

A- 

A- 

4- 
A- 
4' 
4' 
4 
4 
A 
A 
A 
A 



94 

94 

94 

94 

94 

93 

93 
o3 

93 

9 3 

9 3 
92 
92 
.92 
.92 
.92 
.92 
• 91 
.91 
.91 

.91 

9i 

9 1 
90 

90 

90 

90 

90 

90 

89 



4.89 
4.89 
89 
89 
89 
88 
,88 
.88 
,88 



256248 
255900 
200602 
2557 35 
255o57 
204760 
254462 
254i65 
2 53868 
253571 
203274 

10-252977 
202681 

202384 

202087 
201791 
201495 
201 199 
200903 
250607 
25o3n 



10 



10 



IQ 



4-88 

4.88 
4.87 
4-87 
4-87 
4-«7 
4-8i 

4-87 

-i- s '> 
4-86 
4-86 



10 



2500IO 

249719 
249424 
249128 
248833 
248538 
248243 

24794 8 
24-653 
247358 

•247063 
246769 

246474 
246180 
2 p885 
2 4559 1 
243297 
24')oo3 

24 i7°? 
24441 5 

244122 

243-2 4 
24 3535 
243241 
242948 
242635 
242362 
242069 
241776 
2U483 

•24! 190 

240898 

2 4060 5 
24o3l3 
2 i') 121 
7\<r 
239436 
2 So I 14 

a3885a 

23856i 



60 

5 9 

58 
5 7 
56 
55 
54 
53 

52 

5i 
5o 

49 
48 

47 
46 
45 
44 
43 
42 
4i 
40 

3 9 

38 

37 
36 

35 

34 
33 

32 

3i 

3o 

29 
28 

27 
26 

25 

24 

23 

22 
21 
20 

J9 

18 

17 
16 

i5 
14 
i3 
12 
1 1 
10 



CotMllg. 



D. 



Taug. 



7 
6 
5 

4 
3 

2 

1 


M. 



28 



(GO DKGRKER.) 



18 



(30 DEGKEES.j A TABLE OF LOGARITHMIC 



M. j 




Sine 


D. 


Cosine 


D. 


Tang. 


9-698970 


3-64 


9-937531 


I-2I 


9-761439 


i 


699189 


3-64 


937458 


1-22 


761731 


i 


699407 


3-64 


9 37385 


1-22 


762023 


3 


699626 


3-64 


937312 


1-22 


762314 


, 4 


699844 


3-63 


937238 


1-22 


762606 


5 


700062 


3-63 


937165 


I -22 


762897 


6 


700280 


3-63 


937092 


1-22 


763188 


7 


700498 


3-63 


937019 


I -22 


7634-9 


8 


700716 


3-63 


936946 


I -22 


763770 


9 


700933 


3-62 


936872 


1-22 


764061 


10 


7oii5i 


3-62 


936799 


1-22 


764352 


ii 


9.701368 


3-62 


9-936725 


I -22 


9 764643 


12 


701083 


3-62 


936652 


1-23 


764933 


i3 


701802 


3- 61 


936578 


1-23 


765224 


U 


702019 


3-6i 


9365o5 


1-23 


765514 


i5 


to? 286 


3-oi 


9 3643 1 


1-23 


7658o5 


16 


702402 


I- 61 


936357 


1-23 


766095 


17 


702669 


3- 60 


936284 


1-23 


766385 


18 


702885 


3- 60 


936210 


1-23 


766675 


*9 


703101 


3-6o 


936i36 


1-23 


766965 


20 


7o33i7 


3-6o 


936062 


1-23 


767255 


21 


9-7o3533 


3-5 9 


9-935988 


1-23 


9-767545 


22 


703749 


3-5 9 


935qi4 


1-23 


767834 


23 


703964 


3-5 9 


935840 


1-23 


768124 


24 


704179 


3-5 9 


935766 


1-24 


768413 


25 


704395 


3-5 9 


935692 


1-24 


768703 


26 


704610 


3-58 


935618 


1-24 


768992 


27 


704825 


3-58 


935543 


1-24 


769281 


28 


705040 


3-58 


935469 


1-24 


769570 


29 


7o5254 


3-58 


935390 


1-24 


769860 


3o 


705469 


3-5 7 


935320 


1-24 


770148 


3i 


9-7o5683 


3-57 


9-935246 


1-24 


9-770437 


32 


705898 


3-5 7 


935171 


1-24 


770726 


33 


706112 


3-57 


935097 


1-24 


771015 


34 


706326 


3-56 


935022 


1-24 


77i3o3 


35 


706539 


3-56 


934948 


1-24 


771592 


36 


706753 


3-56 


934873 


1-24 


771880 


37 


706967 


3-56 


934798 


1-25 


772168 


38 


707180 


3-55 


934723 


1-25 


77 2 457 


3 9 


707393 


3-55 


934649 


1-25 


772745 


40 


707606 


3-55 


934574 


1-25 


773o33 


41 


9-707819 


3-55 


9-934499 


1-25 


9-773321 


42 


708032 


3-54 


934424 


1-25 


773608 


43 


708245 


3-54 


934349 


1-25 


773896 


44 


708458 


3-54 


934274 


1-25 


774184 


45 


708670 


3-54 


934199 


1-25 


774471 


46 


708882 


3-53 


934123 


1-25 


774759 


47 


709094 


3-53 


934048 


I -25 


775046 


48 


709306 


3-53 


933973 


I -25 


775333 


49 


709518 


3-53 


9 338 9 8 


1.26 


775621 


5o 


709730 


3-53 


933822 


1-26 


775908 


5i 


9.709941 


3-52 


9-933747 


1-26 


9 - -76195 


52 


7ioi53 


3.52 


933671 


1.26 


776482 


53 


710364 


3-52 


933596 


1.26 


776769 


54 


710575 


3-52 


933520 


1.26 


777000 


55 


710786 


3-5i 


933445 


1.26 


777342 


56 


710997 


3-5i 


933369 


1-26 


777628 


ii 


711208 


3-5i 


933293 


1-26 


777915 


711419 


3-5i 


933217 


1-26 


778201 


59 


711629 


3-5o 


933i4i 


1-20 


778487 


60 


711839 


3-5o 


933o66 


1-26 


778774 


Cosine 


! D. 


Sine 
\ 


i I). 


CotlUlGT- 



D. 



4-86 

4-86 
4-86 
4-86 
4-85 
4-85 
4-85 
4 --85 
4-85 
4-85 
4-84 

4-84 

4-84 
4-84 
4-84 
4-84 
4-84 
4-83 
4-83 
4-83 
4-83 

4-83 

4-83 
4-82 
4-82 
4-82 
4-82 
4-82 
4-82 
4-8i 
4-8i 

4-8i 
4-8i 
4-8i 
4-8i 
4-8i 
4-So 
4-8o 
4-So 
4-8o 
4-8o 

4-8o 

4-79 

4-79 
4-79 
4-79 
4-79 
4-79 
4-79 
4-78 
4-78 

4-78 
4-78 
4-78 
4-78 
4-78 
4-77 
4-77 
4-77 
4-77 

4-77 
D. 



Cotang. 



10 



•23856i 
238269 
237977 
23 ;686 
237394 
237103 
2368i2 
236521 
23623o 
235g3o 
235648 

10-235357 
235o67 

234776 
234486 
234195 
233905 
2336 1 5 
233325 
233o35 
232745 

10-232455 
232166 
23i8 7 6 
23 1 587 
231297 
23ioo8 
230719 
23o43o 
23oi4o 

2 2q852 

229563 
229274 
228985 
228697 
228408 
228120 
227832 
227543 
227255 
226967 

•226679 
226392 
226104 
2258i6 
225529 

J2024I 

224954 
224667 
224379 
224092 

io.2238o5 
2235i8 

223231 

222945 

222658 

222372 

222085 

221799 

2?l5l2 

221226 



10 



10 



60 

58 

57 

56 

03 

54 

53 

52 

5i 

5o 

49 
48 

47 
46 
45 
44 
43 
42 
4i 
4o 

3 9 
38 

37 
36 
35 

34 

33 

32 

3i 
3o 

29 
28 

27 
26 

25 

24 

23 
22 
21 

20 

10 
l8 

17 
l6 

i5 

14 
i3 
12 
11 

10 



7 
6 
5 

4 
3 
2 
t 
o 



Tang. I M. 



(f>0 nTCOKF.F.R ") 



SI> T E3 AND TANGENTS. 



(31 



DEGREES.) 



49 



M. 



3 
4 
5 
6 

9 

10 



Sine 



D. 



9-711839 

7I20DO 
712260 
712469 
712679 
712889 
713098 

7i33u8 
7i35i7 



710 



?26 



11 
12 
i3 
14 

ID 

16 

n 

18 

19 
20 

21 
22 

23 

24 

25 

26 

27 
28 

11 

3i 

32 

33 
34 
35 
36 

u 

39 

4o 

4i 
42 
43 
44 
45 
46 

47 
48 

49 
Do 

5i 

52 

53 

54 
55 
56 

57 
58 
5 9 
60 



3-5o 
3 -DO 
3 -do 
3-4o 



713935 

-714144 
714352 
714361 
714769 
714978 
7 1 5 '. 86 

7i5394 
7i56o2 
715809 
716017 

•716224 
716432 
7i663g 
716846 
717053 

^:7259 
717466 
717673 
717879 
71808D 

^•718291 

718497 
718703 
718909 
7191 i4 
719320 
7195^5 
719730 
71993a 
720140 

9-72o3p 
720D49 
7207 
720958 
721162 
72 1 366 
721670 
721774 
721978 
722161 

g-722385 
722588 

722791 

722994 

723197 

723400 

7236o3 

7238o5 

724007 

724210 

Cosmo 



49 
49 
49 
4Q 

48 
48 
48 



3 

q 

>_* 

3- 
3- 
3- 
3- 
3- 

3.43 
3-47 
3-47 
3-47 
3-47 
3-47 
3-46 
3-46 
3-46 
3-46 

3-45 

3-45 

3-45 

3-45 

3-45 

3-44 

3-44 

3-44 

3-44 

3-43 

3-43 

3-43 
3-43 
3-43 
3-42 
3-42 
3-42 
3-42 
3-4i 
3-4i 



Cosine 

9.933066 
932990 

932914 
9 32833 
932762 
9 32635 
932609 
932533 
9324D7 
9 3233o 
9323o4 

9-932228 
932161 

932075 
931998 
931921 
9 3i845 
931768 
93 1 69 1 
931614 
93i537 

931460 
9 3i383 
93i3o6 
931229 
93i i5q 
93107D 
930998 
930921 
930843 
930766 

- 9 3o688 
93061 1 
93o533 
93o456 

930378 
93o3oo 
93o223 
93oi45 
930067 
929989 



D. 



Tan?. 



T>. Cotansr. 



1 -26 

1-27 
1-27 
1-27 

1.27 

1-27 
I -27 
I 
I 
I 



3-4i 

3-4i 

3-40 

3-4o 

3-4o 

3-4o 

3-4o 

3- 3 9 

3-3 9 

3-3 9 

3.39 

3-3 9 
3.33 

3-33 
3-38 
3-33 
3-37 
3-37 
3-37 
3-37 

D. 



)-92Q9TI 

929833 

929755 

929677 

929199 
929521 

92944a 
929364 

929286 

929207 

9-929129 
92oo5o 
928912 
o28 y 9 3 
92881D 
928736 
928657 
9 285 7 8 
928499 
928420 



•27 

•27 

•27 

1-27 

1-27 
1-27 

I-2S 

1-23 

1-28 

1-23 

1.28 
1-28 
1-28 
1-28 

1-28 

1-28 

1.28 

1-29 
I -29 
I -29 
I -29 
1-29 
1-29 
1-29 

1 -29 
1 -29 
1-29 
1-29 
1-29 
i-3o 
i-3o 
i-3o 
i.3o 
i-3o 

i-3o 
i-3o 
1 .3o 
i-3o 
i-3o 
i-3o 
i-3o 
i-3i 
1 -3c 
1 -3i 



i-3i 
i-3i 
1 -3 1 
i-3i 
1 .3i 
i-3i 
1 -3i 
i-3i 
i-3i 
i-3i 



9-778774 
779060 

779 3 46 
779632 
779918 
780203 
7804S9 
78077D 
781060 
781346 
78 1 63 1 

9-781916 
782201 
782486 
782771 
783o56 

783341 
783626 
783910 
784195 
784479 

9-784764 
735o48 
785332 
7856i6 
785900 
786184 
786468 
786752 
787036 
787319 

Q' 78-T6o3 
787886 
788170 
788453 
788736 
789OI9 
789302 
789585 
789868 

7901 5 i 

■790433 
790716 
790999 
791281 
79] 563 
791846 
792128 
792410 
792692 
792974 

) -793 2 56 
793533 
7 9 38ig 
794101 
794383 
79466 \ 
794945 
795227 
795508 

7957 q 9 



4- 
A- 

4- 
4- 
4- 
4- 
4- 
4' 
4' 
4 
4 

4- 
4- 

4- 
4- 
4' 

4' 

4 

4 

4 

4 



77 

77 
76 

76 
76 
76 
76 
76 
76 
'75 
•75 



73 
75 
75 
75 

73 

74 

74 

74 
74 

74 
■74 
•73 
•73 
•73 
•73 
•73 
•73 
•73 



10 



4' 
4' 
4 
4 
4 
4 
4 
4 
4 
4-72 

4-72 
4-72 
4-72 
4-72 
4-72 
4-72 
4-71 
4- 7 i 



7i 
71 



4-7i 

4-7i 



4- 
4' 
4' 
4' 
4- 
4- 
4- 
4- 

4- 
4- 
4- 
4 

4' 
4 
4 



221226 
220940 
220654 
220368 
220082 
219797 
21931 1 
219225 
218940 
2i8654 
?i836 9 

10-218084 
217799 
217314 
217229 
216944 
216659 
216374 
216090 
2i53o5 

2l552I 



Sine 



1). 



Cotantr. 



10 



10 



71 

71 
70 

70 

70 
70 
70 

7° 

70 
69 
69 
.69 

69 
69 
69 



[0 



2i5236 
2i4g52 
21466S 
214334 
214100 
2i33i6 
2i3532 
213248 
212964 
212681 

2 T 2397 
2I2U4 

2ii83o 

2ii547 
21 1264 
210981 
210698 
2 1 04 1 5 

210l32 

209849 

•209567 

209284 

209001 

208719 

208437 

2081 54 

207872 

207 5qo 

2073o8 

207026 



60 
5 9 
58 

57 
56 

55 
54 
53 

52 

5i 

5o 

49 
48 

47 
46 
45 
44 
43 
42 
41 
40 

39 

38 

37 
36 

35 

34 
33 

32 

3i 
3o 

2Q 
20 

27 
26 
25 

24 
23 

22 
21 
20 

IO 
IO 

17 
16 

i5 
14 
i3 

12 
u 
10 



10 



4- 6q 
4-63 
4-68 



200- \ \ 
206463 
206181 
205899 

2o56i7 

2o5336 
2o5o55 
204773 

20 ',492 
2042 1 1 




8 

7 
6 

5 
4 
3 
2 

1 
o 



I). 



Tanft. i M. j 



(58 DEGREES.) 



50 



(32 DEGREES.) A TABLE OF LOGARITHMIC 



M. 
o 


j Sine 


D. 


Cosine 


1 D. 

1 


Tang. 


D. 


i 

Cotaiig. 




7 724210 


3-37 


9 928420 


! 1-32 


9-795789 


4-68 


10-204211 


60 


i 


724412 


3-37 


928342 


I -32 


796070 


4-68 


2o3g3o 




2 


724614 


3-36 


928263 


I -32 


79635i 


4-68 


2o36iQ 


58 


3 


724816 


3-36 


928183 


1-32 


796632 


4-68 


203368 


n 


4 


725o!7 


3-36 


928104 


1-32 


796913 


4-68 


203087 


£ 
j 


J232I9 


3-36 


928025 


j I -32 


797194 


4-68 


202806 


55 


6 


725420 


3-35 


921046 


1-32 


797475 


4-68 


202525 


54 1 


7 


725622 


3-35 


927867 


j 1-32 


797753 


4-68 , 


202245 


1 53 


8 


725823 


3-35 


927787 


1-32 


198036 


4-67 


201964 


5? 


9 


726024 


3-35 


927708 


j 1-32 


798316 


4-67 


20T684 


5i 


10 


726225 


3-35 


927629 


1-32 


798596 


4-67 


20I404 


5o 


it 


o* 72642k 


3-34 


9-927549 


1-32 


9-798817 


4-6 7 


IO-20II23 


% 


12 


726626 


3-34 


927470 


i-33 


799157 


4-67 


200843 


i3 


726827 


3-34 


927390 


1-33 


799437 


4-67 


2oo563 


47 


U 


727027 


3.34 


927310 


i-33 


7997H 


4-67 


200283 


46 


i5 


727228 




927231 


1-33 


799997 


4-66 


20O003 


45 


16 


727428 


3-33 


927151 


i-33 


800277 


4-66 


199723 


44 


H 


727628 


3-33 


927071 


1-33 


800557 


4-66 


199443 


43 


18 


727828 


3-33 


926991 


i-33 


8oo836 


4-66 


199164 


42 


19 


728027 


3-33 


926911 


i-33 


801116 


4-66 


198884 


41 


20 


728227 


3-33 


92683i 


i-33 


801396 


4-66 


198604 


40 


21 


/28427 


3-32 


9 926751 


1-33 


9-801675 


4-66 


10-198325 


3 9 


22 


728626 


3-32 


926671 


1-33 


801955 


4-66 


198045 


38 


23 


728825 


3-32 


926591 


1-33 


802234 


4-65 


197766 


37 


24 


729024 


3-32 


926D1 1 


1-34 


8o?5i3 


A-65 


197487 


36 


2D 


729223 


3-3i 


92643i 


i-34 


802192 


4-65 


197208 


35 


26 


729422 


3-3i 


92635i 


i-34 


8o3o72 


4-65 


196928 


34 


27 


729621 


3-3i 


926270 


1-34 


8o335i 


4-65 


196640 


33 


28 


729820 


3-3i 


926190 


1-34 


8o363o 


4-65 


7 ^/ 

196370 


32 


29 


730018 


3-3o 


926110 


1-34 


803908 


4-65 


196092 


3i 


3o 


730216 


3-3o 


926029 


1-34 


804187 


4-65 


I958i3 


3o 


3i 


• ••73o4i5 


3-3o 


9-925949 


i-?4 
i-^4 


9 • 804466 


4-64 


io- 195534 


29 


32 


73o6i3 


3-3o 


925863 


8o4745 


4-64 


195255 


20 


33 


73o8i 1 


3-3o 


925788 


1-34 


8o5o23 


4-64 


1 9497 7 


27 


34 


73 i 009 


3-29 


925707 


1 -vi4 


8o53o2 


4-64 


194698 


26 


35 


731206 


3-29 


925626 


1-34 


8o558o 


4-64 


194420 


25 


36 


73i4o4 


3-29 


925545 


i-35 


8 585 9 


4-64 


194141 


24 


37 


73t6o2 


3- 29 


925465 


1 -35 


806137 


4-64 


i 9 3863 


23 


38 


731799 


3-29 


925384 


1 -35 


80641 5 


4-63 


193585 


22 


39 


731906 


3-28 


9253o3 


1 -35 


806693 


4'63 


193307 


21 


4o 


732193 


3-28 


925222 


1-35 


806971 


4-63 


193029 


20 


4i 


9-732390 


3-28 


9-925141 


i-35 


9-807249 


4-63 


io- 192751 


m 


42 


732587 


3-28 


925o6o 


1 -35 


807527 


4-63 


192473 


18 


43 


732784 


3-28 


924979 


1 -35 


807805 


4-63 


19:195 


17 


44 


732980 


3-27 


924897 


1 -35 


8o8o83 


4-C3 


19:917 


16 


45 


733177 


3-27 


924816 


1 -35 


8o836i 


4-63 


191639 


i5 


46 


733373 


3-27 


924735 


1-36 


8o8638 


4-62 


191362 


14 


47 


733569 


3.27 


924654 


1-36 


808916 


4-62 


• 191084 


i3 


48 


733760 


3-27 


924572 


i-36 


809 1 93 


4-62 


190807 


12 


4 9 


733961 


3-26 


924491 


i-36 


809471 


4-62 


190529 


II 


5o 


734i57 


3-26 


924409 


1-36 


809748 


4-62 


19025.} 


10 


5i 


9.734353 


3-20 


9-924328 


i-36 


9-810025 


4-62 


io- 189975 


1 


5i 


734549 


3-26 


924246 


i-36 


8io3o2 


4-62 


189698 


53 


734744 


3-25 


924164 


1-36 


8io58o 


4-62 


189420 


7 


54 


73i 9 39 


3-25 


924083 


1-36 


8io85 7 


4 62 


189143 


6 


55 


7 3 5 1 3 5 


3-25 


924001 


1-36 


8m34 


4-6i 


188866 


5 


56 


73:o3o 


1. ~^ 


923919 


1 -36 


811410 


4-6i 


i885qo 


4 


57 


735525 


3- 23 


923837 


1-36 


81 1687 


4-6i 


i883i3 


3 


58 


735719 


3-24 


923755 


i-3 7 


81 1964 


4-6i 


1 880 36 


2 


5 9 


73.'>9i4 


3-24 


9236?3 


I.3J7 


812241 


4-6i 


187759 
i8 7 483 


1 


6o 




736109 


3-24 


923591 


i-3 7 


812517 


4-6i 





i ! 


Cosipe 


I). 


Sine 


D. 


Co&ang. 


D. 


Tang. 


M. 








(57 


DEGRE 


ES.) 









SINES AND TANGENTS (.33 DEGREES.) 



51 



if. 

o 
I 

2 

3 
4 
5 
6 

7 8 

9 
to 

(i 

12 

[3 
14 
i5 
16 

17 

18 

«9 
lo 

21 
22 
*3 

U 

25 
20 

3 

29 

3o 

3i 

32 

33 
34 
35 
36 

37 
38 

3 9 

4o 

4i 
42 
43 
44 
45 
46 

47 
48 

4 9 
5o 

5i 

52 

53 
54 
55 
56 

57 
58 

& 



feme 



D. 



Cosine 



Tang. 



D. 



Cotaner. 



9.736109 
7363o3 , 
736498 
i366o2 
736886 
737080 
737274 
737467 
737661 
737855 
738048 

9-738241 
738434 
738627 
738820 
739013 
739206 
739398 
739090 
739783 
739970 

9.740167 
740339 
74o55o 
740742 
740934 
74H25 
741 3 16 
74i5o8 
741699 
741889 

9-742080 
742271 
742462 
742602 
742642 
743o33 
743223 
7434i3 
7436o2 

743792 

9.743982 
7441-1 
744 501 
744550 
744733 
744928 
745in 
7 i53o6 

745404 
745683 

9-740871 
746109 
746248 
746436 
746624 
746812 
746999 

747187 
747374 
747562 



3-24 
3-24 
3-24 



23 
23 
23 
23 
23 
22 
22 
22 



3 


22 


3- 


22 


8- 


21 


3- 


21 


3- 


21 


3 


21 


3 


21 


3 


20 


3 


20 


3 


20 


3 


20 


3 


20 


3 

3 
3 
3 
3 

3 


19 

19 

19 

-19 

:8 


3 


.18 


3 


• 18 



Cosine 



3-i8 
3-i8 

3-17 
3-17 
3-i 7 
3.17 

3.17 
3-i6 
3-i6 
3- 16 

3- 16 
3- 16 
3 • 1 5 
3-i5 
3 ■ 1 5 
3 • 1 5 
3.. 5 
3-14 
3-i4 
3-14 

3-14 
3 • 1 4 

3 1 3 

4 • 1 3 
3i3 
3- 13 
3- 13 

3-12 
3-12 
3-12 



9-923591 
923509 
923427 
923345 

923263 
923181 
923098 
923016 
922933 
922851 
922768 

9-922686 
922603 
922520 

922438 
922355 
922272 
922189 
922106 
922023 
921940 

9-921807 
921774 

921691 
921607 
921524 
921441 
921357 
921274 

921 190 
921 107 

9-921023 
920939 
920856 
920772 
9206S8 
920604 
920520 
920436 
92o352 
920268 

9-920184 
920099 
92001 5 
919931 
919846 
919762 
919677 
9,9593 
919008 
9194 2 4 

9.919339 

919254 

9191O9 

9 1 90S5 
9 1 9000 
918915 

918830 
918745 
91 HO 59 
918574 



I-3 7 

i-3 7 
l-3 7 
I-3 7 
I-3 7 

i-3 7 
1.37 
i-3 7 
i-3 7 
i-3 7 
1-38 

1-38 
1-38 
1-38 
i-38 
i-38 
1-38 
i-38 
1-38 
1-38 
i-38 



D. 



Sine 



3 9 
3 9 

■ 3 9 

■3 9 

•3 9 
-3 9 
-3 9 
-3 9 
-3 9 
-3 9 

1-39 
1 -4o 
i-4o 
i-4o. 
1 -4o 
1 -4o 
i-4o 
1 -4o 
1 -4o 
i-4o 

1 -4o 
i-4o 
1 -4o 
i-4i 
i-4i 
i-4i 
i-4i 
1 -4i 
i-4i 
i-4i 

1 -4i 

1 -41 
1 -4i 
1 -41 
1-41 

1-42 
1-42 
1-42 
1-42 
1-42 



)-8i25i7 
812794 

813070 

813347 

8i3623 
813899 
8i4i]5 
8i4452 
814728 
8i5oo4 
80279 

3-8i5555 
8i5S3i 
816107 
8 1 6382 
8i6658 
816933 
817209 

817484 
817739 
8i8o3o 

9-8i83io 
8 1 8585 
818860 
8i 9 i35 
819410 
819684 
819959 
820234 
82o5o8 
820783 

9-821057 
82i332 
821606 
821880 
822i54 
82242 
82270 
822977 
82325o 
823524 

9-823798 
824072 
824345 
824619 
82489$ 
825i66 
825439 
825713 
B25986 
826259 

9-826032 
8 2 68o5 
827078 
827331 
827624 
827897 
828170 

8a8442 

828715 

828987 



4-6i 
4-6i 
4-6i 
4-6o 
4- 60 
4-6o 
4- 60 
4-6o 
4- 60 
4-6o 
4-6o 



10 



D. 



5 9 
5 9 
5 9 
5 9 
5 9 

59 
5 9 

5 9 

59 

58 



4-58 
4-58 
4-53 
4-58 
4-58 



58 
58 
58 

57 
57 



4.57 
4.57 
4.57 

4-57 
4.57 
4.57 
407 
4-56 
4-56 
4-56 

4-56 
4-56 
4-56 
4-56 
4-56 
4-56 
4-55 
4-55 
4-55 
4-55 

4-55 
4-55 

4-0) 
4-5'. 
4-55 
4-54 
4 •'">'» 
4 • 54 
4-54 
4-54 



10 



10 



10 



10 



10 



Cot&mr. 



D. 



187482 
187206 
186930 
186653 
1S6377 
1861 01 
185323 
185548 
185272 
184996 
184721 

.184445 
184169 
i83S 9 3 
i836i8 
183342 
183067 
182791 
i825i6 
18224V 
181965 

•181690 
i8i4i5 
1 8 1 1 40 
i8o865 
iSoSgo 
l8o3i6 
1 8004 1 
179766 

179492 
179217 

178943 
178668 

178394 
178120 
177846 
177571 
177297 
177023 
176750 
176476 

176202 
175928 
175655 
I7538l 
175107 
174834 
174'"') 1 
174287 
17401 \ 
173741 

■173468 

173 k/') 
I72022 

172649 
I72376 

17210 5 
171- Jo 

»7i 

17:2s?- 

171013 
Tang. 



60 

57 
56 
55 
54 
53 

52 

5i 

5o 

49 
48 

47 
46 

43 

44 
43 

42 
41 
40 

39 

38 

37 
36 

35 
34 
33 

32 

3i 

3o 

20 

28 

27 
26 

25 

24 
23 
22 
21 
20 

19 
10 

17 
16 

1 5 

14 
i3 
12 
1 1 
10 



7 
6 

5 
4 
3 
2 
1 
o 



(56 DEGREES.) 



52 



(34 DEGREES,) A TABLE OF LOGARITHMIC 



' 


Sine 


D. 


Cosine I 


>. 


Tung. 


P. 


Cotang. 


60 


o 


9' 747562 


3-12 


9-918574 i- 


42 


9-828987 


4-54 


10-171013 


i 


747749 


3 


12 


918489 I- 


42 


829260 


4 


34 


170740 


5 9 


2 


7479 3 6 


3 


12 


918404 1- 


42 


829532 


4 


54 


170468 


58 


4 


748123 


3- 


n 


9 i83i8 i- 


42 


•829805 


4 


54 


170195 


5- 


4 


7483 10 


3- 


1 1 


918233 1 


42 


830077 


4 


54 


169923 


56 


5 


748497 


3- 


11 


018147 1 


42 


83o349 . 


4 


53 


169651 


55 


6 


748683 


3- 


11 


918062 1 


42 


83o62i 


4 


53 


169379 


54 


7 


748870 


3- 


11 


917976 1 


43 


83o8 9 3 


4 


53' 


169107 


53 


8 


749056 


3- 


10 


917891 1 


43 


83n65 


4' 


53 


168835 


5a 


9 


749243 


3- 


10 


917805 1 


43 


83 1437 


4 


53 


168563 


5i 


IO 


749429 


3- 


10 


917719 1 


43 


831709 


4 


53 


168291 


5o 


ii 


9-749615 


3- 


10 


9-917634 1 


43 


9-83i 9 8i 


4 


53 


10-168019 


49 


12 


749801 


3- 


10 


917548 1 


43 


832253 


4 


53 


167747 


48 


i3 


749987 


3- 


09 


917462 1 


43 


832525 


4 


53 


167475 


47 


i4 


750172 


3- 


09 


917376 1 


43 


832796 


4 


53 


167204 


46 


i5 


75o358 


3- 


09 


917290 1 


43 


833o68 


4 


52 


166932 


45 


16 


75o543 


3- 


09 


917204 1 


43 


833339 


4 


52 


1 6666 1 


44 


n 


730729 


3- 


09 


917118 1 


44 


8336n 


4 


52 


i6638 9 


43 


18 


750914 


3- 


08 


917032 1 


44 


833882 


4 


52 


166118 


42 


l 9 


751099 


3- 


08 


916946 1 
916859 i- 


44 


834i54 


4 


52 


165846 


4i 


20 


751284 


3- 


08 


44 


834425 


4 


52 


165575 


4o 


21 


9-751469 


3- 


08 


9-916773 1 


44 


9-834696 


4 


52 


io-i653o4 


39 


22 


75i654 


3- 


08 


916687 1 


44 


834967 


4 


52 


i65o33 


38 


23 


■ 751839 




0- 


08 


916600 1 


44 


835238 


4 


52 


164762 


37 


24 


752023 


3 


07 


9i65i4 1 


44 


835509 


4 


52 


1 6449 1 


36 


25 


752208 


3- 


07 


916427 1 


44 


835780 


4 


5i 


164220 


35 


26 


752392 


3 


07 


9 1 634i 1 


44 


836o5i 


4 


5i 


1^949 


34 


& 


752576 


3- 


07 


916254 1 


44 


836322 


4 


5i 


IL.578 


33 


752760 


3. 


07 


916167 1 


45 


8365 9 3 


4 


5i 


163407 


32 


29 


752944 


3- 


06 


916081 1 


45 


836864 


4 


5i 


I63i36 


3i 


3o 


753128 


3- 


06 


915994 1 


45 


837134 


4 


5i 


162866 


3o 


3i 


9-7533i2 


3 


06 


9-915907 1 


45 


9-8374o5 


4 


5i 


10-162595 


20 


32 


753495 


3 


06 


915820 1 


45 


837675 


4 


5i 


162325 


28 


33 


753679 


3 


06 


9i5733 1 


45 


837946 


4 


5i 


162054 


27 


34 


753862 


3- 


o5 


915646 1 


45 


8382i6 


4 


5i 


161784 


26 


35 


754046 


3 


o5 


9i5559 1 


45 


838487 


4 


5o 


i6i5i3 


25 


36 


754229 


3 


o5 


915472 1 


45 


838757 


4 


5o 


161243 


24 


37 


754412 


3 


o5 


9 i5385 1 


45 


839027 


4 


5o 


160973 


23 


38 


734595 


3 


o5 


915297 1 


45 


839297 


4 


5o 


160703 


22 


3 9 


754778 


3 


04 


915210 1 


45 


83 9 568 


4 


5o 


160432 


21 


40 


754960 


3 


04 


9i5i23 1 


46 


839838 


4 


5o 


160162 


20 


41 


9-755i43 


3 


04' 


9-9i5o35 1 


46 


9-840108 


4 


5o 


10-159892 


10 


42 


755326 


3 


04 


914948 1 


46 


840378 


4 


5o 


159622 


18 


43 


7555o8 


3 


04 


914860 1 


46 


840647 


4 


5o 


i5 9 353 


17 


44 


755690 


3 


04 


9I477 3 J 


46 


840917 


4 


49 


i5 9 o83 


16 


45 


755872 


3 


o3 


914685 I 


46 


841187 


4 


49 


i588i3 


i5 


46 


•j56o54 


3 


o3 


914598 I 


46 


84i457 


4 


49 


158543 


14 


47 


756236 


3 


o3 


9l45lO I 


46 


841726 


4 


49 


1 582 7 4 


1 3 


48 


706418 


3 


o3 


914422 I 


46 


841996 


4 


49 


i58oo4 


12 


49 


736600 


3 


o3 


914334 I 


46 


842266 


4 


49 


157734 


1 1 


5o 


756782 


3 


02 


914246 I 


47 


842535 


4 


49 


157465 


13 


5i 


9-756963 


3 


■ 02 


9«9i4i58 1 


47 


9-842805 


4 


•49 


10-157195 


9 


52 


757144 


3 


• 02 


914070 1 


47 


843074 


4 


•49 


156926 


8 


53 


757326 


3 


■ 02 


913982 1 


47 


843343 


4 


49 


1 5665i 


7 


54 


757507 


3 


•02 


013894 1 


47 


843612 


4 


•49 


156388 


6 


55 


757688 


3 


•01 


9K1006 1 


•47 


843882 


4 


48 


i56ii8 


5 


56 


757869 


3 


•01 


913718 1 


■47 


844i5i 


4 


4S 


1 55849 


4 


u 


758o5o 


3 


•01 


qi363o 1 


■•4 / 


844420 


4 


48 


i555^o 


3 


75823o 


3 


•01 


9i35/, 1 1 


•47 


844689 


4 


48 


i553ii 


2 


59 


75841 1 


3 


•01 


9i3453 1 


•47 


844958 


4 


•48 


i55o42 


1 


60 


758591 


3-oi 


9i3365 1 


•47 


845227 


4-4» 


154773 







Cosine 


P. 


Sine 1 1 


X 


Cotan<r. 


J). 


Tan?. 


M. 



(55 DEGREES.) 



SINES AND TANGENTS. (35 DEGREES.) 



58 



o 
i 

2 

3 
4 
5 
6 

7 
8 

9 

10 

n 

12 

i3 

U 

i5 

16 

i 

i 

19 

20 



Sine 



21 
■22 
23 
24 
25 
26 

27 
28 

3? 

3i 

3 2 

33 

34 

35 

36 

3 

3 

39 
4o 

4i 

42 

43 
44 
45 
46 

a 

5i 

52 

53 
54 
55 
56 

a 

59 

60 



9.758591 

758772 

758932 
739132 
739312 

759492 

759672 
739832 
760031 
76021 1 
760390 

c- 760369 
760748 
760927 
761 106 
761283 

761464 
761642 
761821 
• 761999 
762177 

9.762336 



D. 


3 


01 


3 


■00 


3 


•00 



Cosine 



D. 



Tang. 



D. 



Cotuntr. 



76253.; 

762712 

762889 

763067 

763245 

763422 

763600 

763777 
763934 

.764131 
7 643o3 
764483 
764662 
764838 
765oi5 
765191 
765367 

763344 
765720 

>. 7 658 9 6 
766072 
766247 
766423 
766598 

766774 

7669*9 
767124 
767300 
767473 

9 •767649 
767824 
767999 
768173 
768348 
768322 
768697 
768871 
769045 
769219 



3-oo 
3-oo 
3-oo 
2-99 
2-99 
2-99 
2-99 
2-99 

,98 
.98 

.98 
.98 

.98 
■98 

2-97 
2-97 

2-97 
2-97 

2-97 
2 

2 



9 6 

96 
96 
.96 

. 9 5 

. 9 5 

2-95 
2-95 
2-94 

2-94 
2-94 
94 
94 
94 
93 
93 

9 ^ 
9 3 

•93 

•93 

.92 

.92 

.92 
.92 
.92 
■91 



2-91 

2-91 

2-91 

2-91 

2-90 

2-90 

2-90 

2-90 

2-90 

2-90 



9"9i3365 
913276 
913187 

913099 
9i3oio 
912922 
912833 
912744 
912635 
912566 

912477 

9 . 9 i2388 
912299 
912210 
912121 
9i2o3i 
911942 
91 i833 
911763 
911674 
911384 

9-911495 
9ii4o5 
9 1 1 3 1 5 
911226 
911 i36 
91 1046 
910936 
910866 
910776 
910686 

9-910596 
9io5o6 
9io4i5 
9io325 
9io235 

910U4 
910054 
90996} 
909873 

909782 

• 90969 1 
909601 
909310 

909 i I Q 
909328 

909237 
909 1 16 

900" "> 5 
908964 
908^7 1 

1-908781 
90S690 
908599 
908501 
908416 
908324 
.908333 
908141 
908049 
907938 



Cosine 



D. 



1-47 
i-47 
1-48 

48 

48 

48 

43 

48 
1-48 
1-48 
1-48 

1-48 
1-49 
i-49 
1-49 
1-49 
1-49 

49 

49 

49 

49 

49 

49 
1 -5o 

1 «5o 

1 «5o 

1 -5o 

1 -5o 

1 -5o 

1 -5o 

1 -5a 

1 -5o 
1 >5o 
1 -5o 
i-5i 
i-5i 
i-5i 
i-5i 
i-5i 
i-5i 
i-5i 

i.5i 
i-5i 

i-5i 
i-5i 

1-52 
1-52 
I -f)2 
1 -52 
I -52 
I • 52 

1 .52 

I -32 
I - r )2 
1 52 

1.53 
1.53 
1.53 

1-53 
1-53 
1-53 



Sine 



9-843227 
843496 
845764 
846o33 
846302 
846570 
846839 
847107 
847376 
847644 
847913 

9.848181 
848449 

848717 
8489S6 

849234 
849622 
8^9790 
85oo58 
85o325 
830393 

9-85o86i 
831129 
851396 
85 1 664 
85i93i 
852199 
852466 
852733 
853ooi 
853268 

9-853535 
853802 
854069 
854336 
8546o3 
854870 
855i37 
8554o4 
855671 
855938 

9-8)6204 
856471 
856737 
857004 
837270 
857537 
837803 
838069 
858336 
853602 

9.838868 
8 39 1 34 
859400 
859666 
859932 
860198 

860464 
860730 
860995 
861261 



.. 



4.48 
4-43 
4-48 
4-48 
4-48 
'•47 



47 
47 
47 
47 
47 



4-47 
4-47 



•47 
•47 
•47 



46 

46 



4 

4 

4 

4 

4 

4 

4-46 

4-46 

4.46 
4-46 
4.46 
4.46 
4-46 
4-46 



46 
45 
45 
• 45 



:o 



10 



D. 



Cotang. 



!0 



45 
45 
45 
45 
45 
45 

4-45 
4-45 
4-44 
4-44 

4-44 
4-44 
4-44 
4-44 
4-44 
4-44 
4-44 
4-44 
4-44 
4-43 

4-43 
4.43 
4-43 
4-43 
4-43 
4 • 4 5 
4-43 
4-43 
4-43 
443 



I). 



154773 
i545o4 
i54236 
153967 
1 536 9 8 
1 5343o 
i53i6i 
i528 9 3 
152624 
152356 
152087 

i5i8i9 
i5i55i 
i5i283 
i5ioi4 
150746 
1 5o473 

I302I0 

149942 
149675 
149407 

•149139 
148871 
148604 
148336 
148069 
147801 
147^34 
147267 
146999 
146732 

IO-U6465 
1 46 1 98 
i4593i 

143664 
143397 
i45i3o 
144863 
144^96 
144329 
144062 

10-143796 
143529 
143263 
142996 
142730 
142463 
142 197 

141 
141 398 

I4ii3a 

140866 
1 40600 
1 iiiii 
140068 
139802 
139536 
139270 
139005 

138739 
Tung. 



60 

5q 
58 
5 7 
56 
55 
54 
53 

52 

5i 
5o 

49 
48 

47 
46 
45 
44 
43 
42 
41 
40 

3 9 
38 

37 
36 

35 

34 
33 

32 

3i 
3o 

29 
28 

27 
26 

25 

24 

23 
22 
21 
20 

IO 

18 

17 
16 

i5 
14 
i3 
12 
11 
10 



lio 



7 
6 
5 

4 
3 
2 

1 
o 

M. 



(54 DEGREES.) 



54 



(36 DEGREES.) A TABLE OF LOGARITHMIC 



M. 




Sine 


D. 


Cosine 


D. 


T&ng. 


1 D. 


Cotang. 




9-769219 


2-90 


1 
9-907938 


1-53 


j 9- 861261 


4-43 


10 138739 


! 60 


i 


769393 


2-89 


907866 


1-53 


36 1 527 


4-43 


:38473 


5o 


2 


769666 


2-89 


907774 


1-53 


861792 


4-42 


1 382o8 


58 


3 


769740 


2-89 


907682 


1-53 


862038 


4-42 


137942 


1 5 7 


4 


769913 


2-89 


907590 


1-53 


862323 


4-42 


137677 


56 


5 


770087 


2-89 


907498 


1-53 


86258 9 


4-42 


1 37411 


55 


6 


770260 


2-88 


907406 


1-53 


862854 


4-42 


13-7146 


; 5 4 


I 


770433 


2-88 


907314 


1-54 


863 1 19 


4-42 


1 3688 1 


1 53 


770606 


2-88 


907222 


1-54 


863383 


4-4? 


1 366 1 5 


52 


9 


770779 


2-88 


907129 


1-54 


86365o 


4- 4'* 


1 3635o 


5i 


lo 


7709* 


2-88 


907037 


1-54 


8639i5 


4-4. 


i36oS5 


5o 


II 


9-771125 


2-88 


9-906945 


1-54 


9-864180 


A-Ai 


10-135820 


1 


12 


771298 


2-87 


906852 


1-54 


864445 


4-43 


135555 


i3 


771470 


2-87 


906760 


i-54 


864710 


4 42 


135290 


47 


14 


77i643 


2-87 


906667 


1.54 


864975 


4-4i 


i35o25 


46 


i5 


771815 


2-87 


906575 


1-54 


865240 


4-4i 


134760 


45 


16 


771987 


2-87 


906482 


1-94 


8655o5 


4-4i 


134493 


44 


17 


772159 


2-87 


906389 


1-55 


865770 


4-4i 


i3423o 


43 


18 


77233i 


2-86 


906296 


1-55 


866o35 


4-4i 


133963 


42 


19 


7725o3 


2-86 


906204 


1-55 


8663oo 


4-4i 


133700 


4i 


20 


772675 


2-86 


9061 1 1 


1-55 


866564 


4-41 


133436 


40 


21 


9-772847 


2-86 


9-906018 


1-55 


9-866829 


4-47 


-»• 133171 


3 9 


22 


773018 


2-86 


9o5o25 


1-55 


867094 


4 -4f 


132906 


38 


23 


773190 


2-86 


9o5832 


1-55 


867358 


4-4i 


132642 


37 


24 


77336i 


2-85 


905739 


1-55 


867623 


4-4i 


132377 


36 


25 


773533 


2-85 


9o5645 


1-55 


867887 


4-4i 


i32ii3 


35 


26 


773704 


2-85 


9o5552 


1-55 


868i52 


4-4o 


131848 


34 


27 


773875 


2-85 


9o5459 


1-55 


8684-16 


4-4o 


i3i584 


33 


28 


774046 


2-85 


9o5366 


1-56 


868680 


4-4o 


i3i320 


32 


29 


774217 


2-85 


905272 


1-56 


868945 


4-4o 


i3io55 


3i 


3o 


774388 


2-84 


905179 


1-56 


869209 


4-4o 


130794 


3o 


3i 


9-774558 


2-84 


9-9o5o85 


i-56 


9-869473 


4-4o 


IO.l3o527 

i3o263 


20 


32 


774729 


2-84 


90499 2 


1-56 


869737 


4.40 


20 


33 


774899 


2-84 


904898 


1-56 


870001 


4-4o 


129999 


27 


34 


773070 


2-84 


904804 


i-56 


870265 


4.40 


129730 


26 


35 


775240 


2-84 


9047 1 1 


1-56 


870529 


4-4o 


129471 


25 


36 


775410 


2-83 


904617 


1-56 


870793 


4-40 


129207 


24 


37 


77558o 


2-83 


904523 


1-56 


871037 


4.40 


128943 


23 


38 


775750 


2-83 


904429 


i-5 7 


8 7 i32i 


4.40 


128679 


22 


3 9 


775920 , 


2-83 


904335 


i-5 7 


8 7 i585 


4.40 


I284i5 


21 


4o 


776090 ; 


2-83 


904241 


i.5 7 


871849 


4-3g 


I28i5i 


20 


4i 


9-776259 : 


2-83 


9-904U7 


i-5 7 


9-872112 


4-3 9 


10-127888 


19 


42 


776426 


2-82 


9o4o53 


l-5~ 


872376 


4-3 9 


127624 


10 


43 


7765 9 8 


2-82 


903959 


i-5- 


872640 


4-3g 


127360 


17 


44 


776768 


2-82 


903864 


i-5 7 I 


872903 


4-3 9 


127097 


16 


45 


776937 


2-82 


903770 


i-5 7 


873167 


4-3 9 


126833 


i5 


46 


777106 


2-82 


903676 


i-5 7 


8 7 343o j 


4-3 9 


126570 


14 


47 


777275 


2-8l 


9o358i 


i-5 7 


873694 1 
873967 


4-3 9 


!263o6 


i3 


48 


777444 


2-8l 


903487 


i-5 7 


4-3 9 


126043 ; 


12 


49 


777613 


2-Sl 


903392 


1-58 


874220 


4-3 9 


123780 ; 


ii 


5c 


777781 


2-8l 


903298 


1-58 


874484 


4-3 9 


I255i6 j 


10 


5i 


9.777950 


2-8l 


9-9o32o3 


1-58 


9 '&74747 


4.39 


I0-I25253 


I 


52 


778119 


2- 81 


9o3io8 


1-58 


875010 


4.39 


124990 


53 


778287 


2-80 


9o3oi4 


1-58 


875273 


4>38 


124727 


I 


54 


778455 


2-80 


902919 


1-58 


875536 


4-38 


124464 


55 


778624 


2-80 


902024 


:-58 


875800 


4-38 


124200 


5 


56 


778792 


2-80 


902729 


i-58 


876063 


4-38 


123937 


4 


57 


778960 


2-80 


902634 


1-58 


876326 


4-38 


123674 


3 


58 


779128 


2-80 


902539 


i-5 9 


e-76589 


4-33 


I234H 


2 


^ 


779295 


2-79 


902444 


i-5 9 


876S51 


4-38 


I23l49 


1 


60 


779463 


2-79 


902349 


l.5 9 


877114 


4-38 


122886 







Cc«ine 


D. 


Sine 


D. 


Cotang. 


D. 


Tang-. 


M. 



(53 DEGREES.^ 



SIXES AKD TAXGEXT3. (37 DEGREES.) 



55 



M. 



uJ16 



D. 



Cosine 



D. 



Tang. 



D. 



Coiav.s. 



o 


9.779463 


2-79 


9-902349 


I 


779631 


2 --9 


902253 


2 


7~9~9 S 


2-79 


902153 


3 


7-9966 


2-79 


902063 


4 


-,,33 


2-"9 


9019b- 


5 


- - 3oo 


2-78 


901^-2 


6 


- - ^467 


2-78 


901--6 


1 


7S0634 


2- 7 8 


901681 


8 


780S01 


2-78 


90i5^5 


9 


780963 


2- 7 3 


90UQO 


10 


781134 


2-78 


901394 


ii , 


n -:i3oi 


2-77 


9-901298 


12 


781468 


2-77 


901202 


13 


78l634 


2-TJ 


901 io5 


14 


78 1 800 


2-77 


901010 


i5 


781966 


2-77 


900914 


16 


782-02 


2-77 


90: 


17 


782- 


2-76 


900722 


18 


782 .... 


2-76 


900626 


19 


.7826JO '; 


2-76 


900529 


20 


782- 


2-76 


900433 


21 


9.-S2961 


2.76 


9-900337 


22 


783.27 , 


2-76 


900240 


23 


783292 ! 


2-75 


900144 


24 


7834^8 


2--5 


900047 


23 


783623 


2-75 


8 99 o5 1 


26 

27 
28 


-31- - 
7 83 9 53 
7841 18 


2-75 

2-"5 

2-73 


899804 
899707 
899660 


29 


784282 


2-~4 


899 564 


3o 


784447 


2-74 


899467 


3i 


9-784612 


2-74 


9-8993-0 


32 


78477 


2-"4 


899273 


33 


784941 


2-74 


8991-5 


34 


785io5 


2 --4 


8990-8 

89^981 


35 


- '269 


2-73 


36 


785433 


2-73 


89^ - 4 


37 


785597 


2-73 


898787 


38 


78:- 


2-73 


898689 


3 9 


785920 


2. 7 3 


898092 


40 


786089 


2- 7 3 


898494 


41 


9-786252 


2-72 


i 9-898397 


42 


786416 


2-72 


898299 


43 


- .5-9 


2-2 


89820V 


44 


786-42 


2-71 


89S104 


45 


786906 


2-"2 


898006 


40 


787069 


2-2 


89-908 


47 


- "232 


2-71 


89- 


4^ 


- -395 


2-71 


&97712 


49 


" 55 7 


2-71 


89- . 


DO 


787720 


2-71 


897516 


5i 


9 . 7 8 : 833 


2-71 


9-89-118 


52 


- o45 


2-71 


897320 


53 


--^208 


2-71 


89-222 


54 


7883-;o 


2-70 


89 •'123 


55 


7 R8532 


2-70 


- 


56 


7« 


2-70 


89 


ll 


788806 


2-70 


5828 


789018 


2 70 


896-29 


59 


789180 


2--0 


663 1 


60 


789342 


269 


896532 


Cosine 


D. 


Sine 



09 
09 
09 
09 

I -09 

I 

I 

I 

I 

I 



09 
09 
09 
09 
.09 

1 -6o 

1 -6o 
1 -6o 
1 -6o 
1 -6o 
I -6o 
I- 60 
1 -6o 
1 -6o 
I- 60 
1.6] 



9-877114 

877377 
8-7640 
S--903 
878 i65 
--428 
8-8691 
-> 9 53 
8-9216 
8-9478 
879^41 

g-88ooo3 
88o265 
8800? 3 
880-90 
881062 
88i3i4 
88i5-6 
881839 
882:^1 
882363 



■64 9 

■64 
-64 
.64 
■64 
•64 
■64 
■64 
•64 
• 64 



4-33 
4-33 
4-33 
4-33 



10 



i-6i J 9.882625 

i- 61 882887 

1 -6i 883U8 ! 

i-6i 833410 ! 

i-6i ; 883672 

1-6 1 833934 i 

i-6i 884196 I 

i-6i 884407 I 

i-6i 884-19 

1-62 884980 

1-62 9-885242 

1-62 8355o3 

1-62 885765 

1-62 8S6026 

1-62 886288 

1-62 886549 

1-62 886810 

1-62 887072 

1-62 83-333 

1-63 887394 

1-63 9 -887855 

1-63 -116 

1-63 8*33-7 

1-63 888639 

63 -900 

•63 889160 

•63 .',21 

• 63 J682 

•63 8^99 j3 
•63 1 890204 



890465 
890-20 
8909S6 

891247 
891007 
H91- - 
892028 
8922S9 
892049 
S92810 



33 
33 
33 

37 
37 
37 
3 7 



4-3 7 
4-3 7 
4-3 7 
4-3 7 
4-37 
4-3 7 
4 -37 
4-37 
4-37 
4-36 

4-36 

4-36 
4-36 
4-36 
4-36 
4-36 
4-36 
4-36 
4-36 
4-36 

4-36 
4-36 
4-36 
4-36 
4-36 
4-35 
4-35 
4-35 
4-35 
4-35 



35 
35 
35 
35 
35 
35 
4-35 
4-35 
4-35 
4-34 

4-34 
4-34 
4-34 



34 
34 
34 
34 
4-34 
4-34 
4-34 



IC' 



122880 
122623 

122360 

122097 
1 2 1 S35 
1 2 1 5-2 
121809 
1 2 1 047 
120784 
120022 
120209 

119997 

119-30 
1194-2 
119210 
118948 
1 1 8686 . 
1 1 8424 : 
I1S161 
11-899 ' 
1 17637 1 

io-ii "3-5 
1 171 i3 

ii6S52 
1 16590 

116328 

1 1 6066 
n53o4 
iio543 
ii523i 
110020 

10-114-53 

1 14497 
1 14235 

1 139-4 
1 1 3- 1 2 

ii34oi 
1 1 3190 
112923 
1 1 2667 
1 1 2406 

io- 112140 
111 

1 n623 
iii36i 
1 1 1 1 00 
ii' 

1 1 - 
no 

in " 
1 09 - 

io- 1 09030 
109275 

109014 
10- 

108 

10M232 

10" 

10--1 1 

10- 

107190 



60 

5o 

53 

57 
56 
55 
54 
53 

52 

01 
5o 



40 

47 
46 
45 
44 
43 
42 
4i 
40 

3 9 
33 

37 
35 
35 
34 
33 

•32 

3i 
3o 



27 
26 

20 

24 

23 
22 
21 
20 

IO 
13 

17 

16 
10 
14 

i3 
12 
11 
10 



D. 



Cotaiig. 



D. 



Tau 



7 
6 
5 
4 
3 

3 
I 
O 

H. 



(52 DEGREES.) 



56 



(38 DEGKEES.) A TABLE OF LOGARITHMIC 



M. 




Sine 


D. 


Cosine I 


). 


Tang. 


D. 


Cotang. 




9*789342 


2-69 


9-896532 1 


• 64 


9-892810 


4-34 


10-107190 


60 


i 


789504 


2 


.69 


896433 1 


-65 


893070 


4 


-34 


106930 


5o 


2 


789665 


2 


.69 


8 9 6335 1 


-65 


8 9 333 1 


4 


-34 


106669 


58 


3 


789827 


2 


•69 


896236 1 


-65 


893591 


4 


-34 


106409 


57 


4 


789988 


2 


•69 


896137 1 
8 9 6o38 1 


-65 


8 9 385i 


4 


-34 


106 149 


56 


5 


790149 


2 


•69 


•65 


8941 1 1 


4 


-34 


io588 9 


55 


6 


7903 10 


2 


• 68 


8 9 5o39 1 


-65 


894371 


4 


• 34 


I0562Q 


54 


7 


790471 


2 


• 68 


895840 1 


-65 


894632 


4 


• 33 


io536o 


53 


8 


790632 


2 


• 68 


895741 1 


• 65 


894892 


4 


• 33 


io5io8 


52 


9 


790793 


2 


• 68 


895641 1 


-65 


895162 


4 


■33 


104848 


5i 


10 


790954 


2 


■ 68 


895542 1 


•65 


895412 


4 


•33 


104588 


5o 


ii 


9-791115 


2 


• 68 


9-895443 1 


•66 


9-895672 


4 


■ 33 


10-104328 


4o 


12 


791275 


2 


.67 


895343 1 


• 66 


89D932 


4 


•33 


104068 


48 


i3 


791436 


2 


.67 


895244 1 


• 66 


896192 


4 


•33 


io38o8 


47 


14 


79i5o6 


2 


.67 


895145 1 


■ 66 


896452 


4 


-33 


io3548 


46 


i5 


791767 


2 


.67 


895045 1 


■ 66 


896712 


4 


-33 


103288 


45 


16 


791917 


2 


•67 


894945 1 


■ 66 


896971 


4 


-33 


103029 


44 


17 


792077 


2 


.67 


894846 1 


66 


897231 


4 


-33 


102769 


43 


18 


792237 


2 


66 


894746 1 


66 


897491 


4 


-33 


102009 


42 


19 


792397 


2 


66 


894646 1 


66 


897751 


4 


•33 


102249 


4i 


20 


792557 


2 


66 


894546 1 


66 


898010 


4 


• 33 


101990 


40 


21 


9-792716 


2 


66 


9-894446 1 


67 


9-898270 


4 


■33 


10-101730 


3o 


22 


792876 


2 


66 


894346 1 


67 


8 9 853o 


4 


• 33 


101470 


38 


23 


793o35 


2 


66 


894246 1 


67 


898789 


4 


33 


101211 


37 


24 


793io5 


2 


65 


894146 1 


67 


899049 


4 


■32 


1 0095 1 


36 


25 


793354 


2 


65 


894046 1 


67 


899308 


4 


■32 


100692 


35 


26 


7935 1 4 


2 


65 


893946 1 


67 


899568 


4 


■32 


ioo4o2 


34 


27 


793673 


2 


65 


893846 1 


67 


899827 


4 


•32 


1 001 73 


33 


1 28 


793832 


2 


65 


893745 1 


67 


900086 


4 


32 


099914 


3a 


9 


793991 


2 


65 


893645 1 


67 


900346 


4 


32 


099654 


3i 


1 3o 


7941 5o 


2 


64 


893544 1 


67 


900605 


4 


32 


099395 


3o 


1 3i 


9» 7943o8 


2 


64 


9-893444 1 


68 


9.900864 


4 


32 


io-o90i36 


29 


1 ll 


794467 


2 


64 


893343 1 


68 


901124 


4 


32 


09^876 


28 


33 


794626 


2 


64 


893243 1 


68 


90i383 


4 


■32 


098617 


2 7 


34 


794784 


2 


64 


893142 1 


68 


901642 


4 


•32 


098358 


26 


35 


794942 


2 


64 


893041 1 


68 


901901 


4 


32 


098099 


25 


36 


79D101 


2 


64 


892940 1 


68 


902160 


4 


32 


097840 


24 


37 


795259 


2 


63 


892839 i- 


68 


902419 


4 


32 


097081 


23 


38 


795417 


2 


63 


892739 1 


68 


902679 


4 


32 


097321 


22 


3 9 


795575 


2- 


63 


8 9 2638 1. 


68 


902938 


4 


32 


097062 


21 


40 


795733 


2- 


63 


8 9 2536 1. 


68 


903197 


4 


3i 


096803 


20 


4i 


9-795891 


2- 


63 


9-892435 i- 


69 


9-903455 


4 


3i 


10-096545 


IQ 


42 


796049 


2- 


63 


892334 I- 


69 


903714 


4 


3i 


096286 


l8 


43 


796206 


2- 


63 


892233 1. 


69 


903973 


4 


3i 


096027 


17 


44 


796364 


2- 


62 


892132 i- 


69 


904232 


4 


3i 


090768 


16 


45 


796521 


2- 


62 


892030 1 • 


69 


904491 


4 


3i 


095509 


i5 


46 


796679 


2- 


62 


891029 i- 


69 


904700 


4 


3i 


095250 


14 


% 


796836 


2- 


62 


891827 i- 


69 


905008 


4 


3i 


09499 2 


i3 


796993 


2- 


62 


891726 i- 


69 


905267 


4 


3i 


094733 


12 


49 


797100 


2- 


61 


891624 i- 


69 


905526 


4 


3i 


094474 


11 


5o 


797307 


2- 


61 


891523 i- 


70 


905784 


4 


3i 


094216 


10 


5i 


9.797464 


2- 


61 


9-891421 i- 


70 


9-906043 


A- 


3i 


10-093957 


9 


52 


797621 


2- 


61 


8913 19 i- 


70 


906302 


4' 


3i 


093698 





53 


797777 


2- 


61 


89 1 2 1 7 1 • 


70 


9o656o 


4 


3i 


093440 


7 


54 


797934 
798091 


2 


61 


891 1 1 5 i- 


70 


9068 1Q 


A- 


3i 


093181 


6 


55 


2 


61 


891013 i- 


70 


907077 


A- 


3i 


092923 


5 


56 


798247 
79840J 


2 


61 


89091 1 i- 


70 


907336 


A- 


3i 


092664 


4 


5 7 


2 


60 


890809 1 • 


70 


907594 


A- 


3i 


092406 


3 


58 


798560 


2 


60 


890707 1 • 


70 


907802 
9081 1 1 


4- 


3i 


092 1 48 


2 


5 9 


798716 


2 


60 


8qo6o5 1 • 


70 


4- 


3o 


091889 


1 


60 


798872 


2-6o 


890D03 1 • 


70 


908369 


4-3o 


091631 





1 . 


Co«ine 

— ■ ... — 1 ■ , — ■ 


— ■ .i — - 1 ■■ 


Sine L 


». 


Cotang. 


D. 


Tai:g. 


M. 



(51 DEGREES.) 





SINES AND TANGENTS. 


(39 DE 


;gr 


liiiiiO( 


■ M 







Sine 


D. j 


Cosine D 


, 


Tang. ! 


D. 


Cotang. 


1 
60 j 


9-798872 


2- 60 


9-890503 i- 


70 


9-908369 


4-3o 


10-091 63 1 


i ! 


o 

799C20 


2- 


60 


890400 1 • 


7i 


008628 


4- 


60 


091372 


5o 


2 


799184 


2- 


60 


890298 1 • 


71 


908886 


4- 


3o 


091 1 14 


e 


3 


799339 


2- 


09 


890195 i- 


7i 


909144 


4- 


3o 


090806 


57. 


4 


799490 


2- 


5 9 


890093 1 • 


7i 


909402 


4- 


3o 


090098 


56 


5 


799601 


2- 


5 9 


889990 1 • 


7i 


909660 


4- 


3o 


090340 


5o 

C t 


e 


799806 


2- 


5 9 


889838 1 • 


7i 


90Q918 


4 


3o 


090082 


54 


i 


799962 


2- 


5 9 


889785 1 • 


7i 


910177 


4' 


3o 


089823 


53 


8001 17 


2- 


59 


889682 1 


7i 


910430 


4' 


3o 


089060 


52 


Q 


800272 


2- 


58 


889079 1 


7i 


910693 


4 


3o 


0S9307 


5i 


10 


800427 


2- 


58 


889477 1 


7i 


910901 


4 


3o 


089049 


5o 


II 


9-8oo582 


1- 


58 


9-889314 1 


72 


9-91 1209 


4 


3o 


10-088791 


4Q 


12 


800737 


2 


58 


889271 1 


72 


911467 


4 


3o 


o88533 


48 


lJ 


800892 


2 


58 


889168 1 


72 


911724 


4 


3o 


088276 


47 


14 


801047 


2 


58 


889064 1 


72 


9119^2 


4 


3o 


088018 


46 


15 


801201 


2 


58 


888961 1 


72 


91 2240 


4 


3o 


087760 


40 


16 


8oi356 


2 


57 


888808 1 


72 


912498 


4 


3o 


087502 


44 


» i 


8ui5n 


2 


5 7 


888705 1 


72 


912706 


4 


Jo 


087244 


43 


18 


801 665 


2 


57 


88865 1 1 


72 


9i3oi4 


4 


29 


086986 


42 


J 9 


' 801819 


2 


57 


888548 1 


72 


913271 


4 


29 , 


086729 


4i 


20 


801973 


2 


57 


888444 1 


73 


913529 


4 


29 


08647 1 


40 


21 


9-802128 


2- 


57 


9-888341 1 


73 


9-913787 


4 


29 


io-o862i3 


In 


22 


802282 


2- 


56 


888237 1 


73 


914044 


4 


29 


085906 


38 


23 


802436 


2- 


56 


888i34 1 


73 


9i43o2 


4 


29 


080698 


37 


24 


802089 


2- 


56 


888o3o 1 


73 


914060 


4 


29 


o8544o 


36 


25 


802743 


2- 


56 


887926 1 


73 


914817 


4 


29 


o85i83 


35 


26 


802897 


2- 


56 


887822 1 


73 


9i5oi5 


4 


29 


084925 


34 


2 7 


8o3ooo 


2 


56 


887718 1 


73 


915332 


4 


29 


084668 


33 


28 


803204 


2 


56 


887614 1 


73 


90090 


4 


29 


084410 


32 


2 9 


8o3357 


2 


55 


887510 1 


73 


9 > 5847 


4 


29 


o84i53 


3i 


3o 


8o3ou 


2 


55 


887406 1 


74 


9 1 6 1 o4 


4 


29 


083896 


3o 


3i 


9 -8o3664 


2 


55 


9-887302 1 


74 


9-916362 


4 


29 


io-o83638 


20 


32 


8o38i7 


2 


55 


887198 1 


74 


9 1 66 1 9 


4 


29 


o8338i 


28 


33 


803970 


2 


55 


887093 1 


74 


9.6877 


4 


29 


o83i23 


27 


34 


804123 


2 


55 


886989 1 
886885 1 


74 


9«7'34 


4 


29 


082866 


26 


35 


804276 


2 


54 


74 


91739' 


4 


29 


082609 


25 


36 


804428 


2 


54 


886780 1 


74 


917648 


4 


29 


082302 


24 


ll 


804081 


2 


54 


886676 1 


74 


917905 


4 


20 


082095 


23 


8o4734 


2 


54 


886571 1 


'74 


9i8i63 


4 


• 28 


081837 


22 


3 9 


804886 


2 


54 


886466 1 


■74 


918420 


4 


28 


o8i58o 


21 


4o 


8o5o39 


2 


54 


886362 1 


•75 


918677 


4 


.28 


o8i323 


20 


4i 


9-805191 


2 


54 


9. 88625 7 1 


■75 


9-918934 


4 


.28 


10-081066 


19 


42 


8o5343 


2 


53 


8861 52 1 


■7* 


919191 


4 


.28 


080809 


10 


43 


800490 


2 


53 


886047 1 


•75 


919448 


4 


.28 


o8o552 


17 


44 


800647 


2 


53 


880942 1 


•75 


919705 


4 


■ 28 


080295 


16 


45 


800799 


2 


• 53 


88583 7 1 


•75 


919962 


4 


■ 28 


o8oo33 


i5 


46 


805901 


2 


• 53 


885732 1 


•75 


920219 


4 


• 28 


079781 


14 


a 


806 io3 


2 


• 53 


885627 1 


•7 5 


920476 


4 


• 28 


079524 


i3 


806204 


2 


■ 53 


885522 1 1 


■75 


920733 


4 


■ 28 


079267 


12 


4o 


806406 


2 


•52 


880416 1 


•75 


920990 


4 


.28 


079010 


11 


5o 


8o65o7 


2 


•52 


8853 1 1 1 


.76 


921247 


4 


-23 


078703 


10 


5i 


0-806709 


2 


•52 


9-885205 1 


.76 


9-92i5o3 


4 


.28 


10-078497 


% 


52 


806860 


2 


-52 


885ioo 1 


.76 


921760 


4 


28 


078240 


53 


80701 1 


2 


•52 


884994 1 
884889 1 
884783 1 


.76 


922017 


4 


• 28 


0779 s3 


I 


54 


807163 


2 


•52 


.76 


922274 


4 


• 28 


077726 


55 


807314 


2 


•52 


.76 


922030 


4 


.98 


077470 


5 


56 


807465 


2 


-5i 


884677 1 


.76 


922787 


4 


• 28 


077213 


4 


£ 


807615 


2 


• 5i 


88/,5 7 2 1 


.76 


923644 


4 


• 28 


076956 


3 


807766 


2 


•5i 


884466 1 


.76 


923300 


4 


•23 


076700 


2 


5g 


807917 
808067 


2 


-5i 


884360 1 


.76 


923557 
9 238i3 


4 


•27 


076/, 1 3 


1 


6o 


2-5l 


884254 1 


•77 


4-27 


076187 



M. 




Cosine 


D. 


Sine I 


). 


Cotang. 


D. 


Tang. 



(50 DEGKEES.) 



58 



(40 DEGREES.) A TABLE OF LOGARITHMIC 



M. 




Sine 


D. 


Cosine 


D. 


Tang. 


D. 


Cotang. 




9-808067 


2-5l 


9.884254 


1-77 


9-9238i3 


4-27 


10 076187 


60 


i 


808218 


2-5l 


884148 


i-77 


924070 


4-27 


075930 


5o 


2 


8o8368 


2-5l 


884042 


1-77 


924327 


4-27 


075673 


I 58 


3 


808519 


2-5o 


883 9 36 


1-77 


92 4583 


4-27 


075417 


1 5 7 


4 


808669 


2-5o 


883829 


1-77 


924840 


4-27 


075160 


56 


5 


808819 


2-5o 


883723 


1-77 


920096 


4-27 


074904 


55 


6 


808969 


2-5o 


883617 


i-77 


920302 


4-27 


074648 


; 54 


7 


8091 19 


2-5o 


8835io 


1-77 


925609 


4-27 


074391 


53 


8 


809269 


2-5o 


883404 


I'll 


925865 


4-27 


074135 


52 


9 


809419 


2-49 


883297 


1.78 


926122 


4-2 7 


073878 


5i 


10 


809369 


2-49 


883191 


I.78 


926378 


4-27 


073622 


5o 


II 


9-8c;7i8 


2-49 


9-883084 


1.78 


, 9-926634 


4-27 


10-073366 


49 


12 


8,9868 


2-49 


882077 

882871 


1.78 


j 926890 


4-2 7 


073 1 10 


48 


i3 


& 100 1 7 


2-49 


1.78 


927'47 


4-27 


072803 


47 


14 


810167 


2-49 


882764 


1.78 


927403 


4-27 


072097 


46 


15 


8io3i6 


2-48 


882607 


1-78 


927639 


4-27 


072341 


45 


16 


8io465 


2-48 


88255o 


1.78 


9279:5 


4-27 


072085 


44 


17 


810614 


2-48 


882443 


1.78 


928171 


4-27 


071829 


43 


18 


810763 


2-48 


882336 


1.79 


928427 


4-2 7 


071573 


42 


IQ 


810912 


2-48 


882229 


1-79 


928683 


4-27 


07 i3i 7 


4i 


20 


811061 


2-48 


882121 


!-79 


928940 


4-27 


071060 


40 


21 


9-81 1210 


2-48 


9-882014 


1-79 


9-929196 


4-2 7 


10-070804 


39 


22 


8n3o8 


2-47 


881907 


1.79 


929402 


4-2 7 


070048 


38 


23 


81 > 007 


2-47 


88. 799 


1.79 


92970S 


4-2 7 


070292 


ll 


24 


8u655 


2-47 


881692 


1.79 


929964 


4-26 


070036 


25 


Pi 1804 


2-47 


881 584 


1-79 


930220 


4-26 


069780 


35 


26 


811952 


2-47 


881477 


1.79 


93o475 


4-26 


069525 


34 


27 


81 2100 


2-47 


881 36 9 


1.79 


930731 


4-26 


069269 
o6qoi3 


33 


28 


812248 


2-47 


8S1261 


i-8o 


930987 


4-26 


32 


29 


8i23 9 6 


2-46 


88n53 


i-8o 


93i243 


4-26 


068757 


3i 


3o 


812344 


2-46 


881046 


i-8o 


931499 


4-26 


o685oi 


3o 


3i 


• 8 1 2692 


2-46 


9-880938 


i-8o 


9-931755 


4-26 


10-068245 


29 


32 


812840 


2-46 


88o83o 


i-8o 


932010 


4-26 


067990 
067704 


28 


33 


812988 


2-46 


880722 


i-8o 


932266 


4-26 


27 


34 


8i3i35 


2-46 


88o6i3 


i-8o 


932522 


4-26 


067478 


26 


35 


8i3283 


2-46 


88o5o5 


i-8o 


932778 


4-26 


067222 


25 


36 


8i343o 


2-45 


880397 


i-8o 


933o33 


4-26 


066967 


24 


•3 7 


8i35 7 8 


2-45 


880289 


i-8i 


933289 


4-26 


0667 1 1 


23 


38 


813725 


2-45 


880180 


1 -81 


933543 


4-26 


066400 


22 


3q 1 


8i38 7 2 


2-45 


880072 


1-81 


9338oo 


4-26 


066200 


21 


4o 


814019 


2-45 


879963 


1-81 


934o56 


4-26 


o65g44 


20 


4i 


9-814166 


2-45 


9-87q855 


i-8i 


9-9343i 1 


4-26 


io- 06 568o 
o6o433 


IO 


42 


8u3i3 


2-45 


879146 


1*81 


934567 


4-26 


10 


43 


814460 


2-44 


8796J7 


1-S1 


934823 


4-26 


065177 


17 


44 


814607 


2-44 


879329 


i-8i 


930078 


4-26 


064922 


16 


45 


8i4753 


2-44 


879420 


i-Si 


935333 


4-26 


064667 


i5 


46 


8i4qoo 


2-44 


8793 1 1 


1 81 


930089 


4-26 


0644 1 1 


14 


47 


810046 


2-44 


87920? 


1-82 


935844 


4-26 


064136 


i3 


48 


8:5i 9 3 


2-44 


879093 1 


1-82 


936100 


4-26 


063900 


12 


49 


8i5339 


2-44 


878984 


I 82 


936355 


4-26 


063645 


11 


5o 


8 1 5483 


2-43 


878875 


1-82 


936610 


4-26 


063390 


10 


5i 


9 -8i563i 


2-43 


9-878766 


1-82 


0-936866 


4-25 


io-o63i34 


9 


52 


815778 


2-43 


878656 


1-83 


937121 


4-25 


062879 


8 


53 


815924 


2-43 


878547 


1-82 


937076 


4-25 


062624 


I 


54 


816069 


2-43 


878438 


1-82 


937S32 


4-25 


062368 


55 


816216 


2-43 


878328 


1-82 


q3 7887 


4-25 


0621 1 3 


5 


56 


8i636i 


2-43 


878219 


1-83 


. 938145 


4-25 


o6i858 


4 


n 


8i65o7 


2-42 


878109 


1-83 


93839b 1 


4-25 


061602 


3 


8i6652 


2*42 


877999 


i-83 


93860$ 1 


4-25 


06 1 347 


2 


59 


8167^3 


2-42 


877890 


1-83 


9 38 9 o8 ' 


4-2i 


06109'' 
o6o83i 


1 


6o 


816943 


2-42 


877780 


1-83 


939163 1 


4-a-* 

1 







Coeine 


D. 


Sine 1 


D. 


Cotang. 


D. , 


Tang. 


M. 



(49 DEGREES.) 



SINES AND TANGENTS. (41 DEGREES.) 



69 



M. 



o 

i 

2 

3 

4 

5 
6 



Sine 



9 

xO 

ii 

12 

i3 
14 
i5 
16 

17 
18 

19 

20 

21 
22 
23 
24 
23 
26 

27 
28 

29 

3o 
3i 

32 

33 
34 
35 
36 

37 
38 

3 9 
4o 

41 

42 

43 

44 

45 

46 

4 

4 

it 

5i 

52 

5J 
54 
55 
56 

n 



D. 



60 



9.816943 
817088 
817233 

817379 

81-524 
817668 
8l73l3 
817938 
8i«io3 

818247 

81S392 

9 .8i8536 
818681 
818825 

8 1 8969 
8191 i3 
819257 
819401 
.819345 
819689 
819832 

9-819976 
820120 
820263 
820406 
82o55o 
820693 
8208J6 
820979 
821 122 
821265 

)■ 82 1 407 
8ji55o 
821693 
82l8J5 

821977 
822120 
822262 

822404 

822346 

822688 

9.822H30 
822972 
823114 
823255 
823307 

82 3)39 
82 J680 

823^21 

82 5963 
824104 

9-824245 

824I86 
824)27 
824668 
824808 

824949 

82 )o<;o 

82 ')2 io 
82537I 

8255i 1 



42 
42 
42 

42 

41 
41 
41 

•4: 
•41 
■41 
•41 



2-40 
2-40 



40 

40 
40 
40 
40 

3g 

39 
39 

39 

39 

3 9 

39 
2-33 

2-38 

2-33 

2-38 

2-38 

2-33 

2.38 
2.38 

2.3 7 

2-3 7 

2-3 7 

2-3 7 

2.37 

2-3 7 
2 .37 

2-36 

2-36 
2 • 36 
2-36 
2-36 
2-36 
2-36 
2-35 
2-35 
2-35 
2-35 



2 


35 


2 


35 1 


2 


35 


2 


34 


2 


34 


2 


•34 


2 


•34 


2 


•3', 


2 


•34 


2 


•34 



Cosine 

-377780 
871670 
877360 
8774-0 
877340 
87723o 
877120 
87701c 
876899 
876789 
876678 

-876568 
876437 

876347 
876236 
876:25 
876014 
873904 

873793 
875682 
875571 

1-875439 
875348 
875237 
875126 
875014 
8749°3 

87479' 

874680 

874303 
874456 

)■ 874344 
874232 
8741 2 1 
874009 
873H96 
873784 
873672 
873360 
873448 
873335 

9-873223 
873110 

8-2998 
872883 
872772 
8726 >') 
872347 
872434 
872821 
872208 

9-872095 
B71981 

871H03 
871755 
87 1 6 ', 1 
871528 
871/,1/i 
871 5»i 
871181 
871073 



D. 

• 83 
1-83 
1-83 
i-83 
1-83 
1-84 
1-84 
1-84 
1-&4 
1-84 
1-84 

1-84 

1-84 
1-84 
1-85 
1-85 
1-85 
1-85 
1-85 
1-85 
1-85 

1.85 
1-85 
1-85 
1.86 
1-86 
1.86 
1-86 
1-86 
1.86 
1-86 

1-86 
1-87 
1.87 
1-87 
1.87 
..87 
..87 
I.87 
1-87 
1-87 

I • ss 
1.83 
1.88 
,.88 
1 ■ 
E .88 
,,83 
1.88 
i-83 

1-89 
I.89 
1.89 
1 .89 
1-89 
I.89 
I.89 
I-8q 
I- 89 
I • 90 



Tans. 



< Joaine 



D. 



Sine 



D. 



Cotaiur. 



9-939163 

989418 
939673 
989928 
94oi83 
940438 
940694 
Q40Q49 
941204 
941433 
9417U 

9-941 968 
942223 
942478 
942733 
942988 
943243 
943498 
943732 
944007 
944262 

9.944517 

944771 
945026 
943281 
945535 

943790 
946045 
946299 
946554 
946808 

9-947063 
9473i8 

947372 
947826 
948081 

9 ',8336 
918590 

9 i8844 

949099 
9493 5 3 

9-949607 
949862 
9 5oi 16 
950370 
900625 



9)1 1 3, 
951888 
951642 
951896 

•952i5o 
9524o5 
93265o 
95291 J 
93 J 1 67 
953421 
9") i'i-5 
9 53 

9 54i83 
934437 



25 

25 

25 

23 

25 

25 

•25 

23 

•25 

•25 

•25 



4-25 

4 25 

4- 

4- 

4« 

4- 

A- 

4' 

4' 

4 



A- 
4- 
4- 
4- 
4- 
4- 
4' 
4' 
4 
4 

4 
4 
4 
4 
4 
4 
4 
4 
4 



I). 



Cotang. 



23 

25 
25 
25 
25 
25 
25 
25 

25 

24 
24 
24 

24 

24 
24 
24 
24 
•24 

•24 
•24 
•24 
•24 
•24 
•24 
•24 
•24 
•24 
•24 



4- 

4- 
4- 
4- 



24 

24 
24 
24 
4-24 
4-24 
4-24 
4-24 
4-24 
4-24 

4«*4 

4-24 
4-24 
4-24 
4-23 
4 23 
4- 23 
4-23 
4-2 J 
4 t8 



10-060837 
o6o582 
060327 
060072 
059817 
059562 
059306 
059051 
058796 
058542 
058286 

io-o58o32 

057777 
057522 
057267 
057012 
056757 

036302 

056248 
055998 
055733 

10-055433 
055229 
054974 

054719 
054460 
054210 
053955 
o537<"i 
o5344v 
053192 

io-o52l8i 
0526S2 
052428 
052174 
051919 
o5i664 
o5i4io 
o5 1 1 56 
50901 
050647 

10-050393 
o5oi J8 
049884 
049630 
049375 
049 1 2 1 
048867 
048612 
o;M")8 

04S104 

10-047850 
04759D 

o i- ] 1 1 

04- 

046,13 
04' 
o ,'■ 
041 1 
04 

l>4 ' Vl > 



D 



60 
5 9 
58 

57 
56 
55 

54 
53 

52 

5i 

5o 

49 
48 

47 
46 

43 

44 

43 
42 
41 
40 

39 
38 

37 
36 
35 

34 
33 

32 

3i 
3o 

29 
28 

27 
26 

25 

24 

23 
22 
21 

20 

IO 
IO 

>7 
16 
i5 
14 
1 3 
12 
11 
10 



_2^L 



7 

6 
5 

4 
3 
2 

1 
o 

M. 



(48 DEGREES.) 



60 



(42 DEGREES.) A TABLE OF LO&ARITHinO 



M. 




Sine 


D. 


Cosine D. 


Tang. 


D. 


Cotang. 




9-8255ii 


2-34 


9-871073 1 


.90 


9-954437 


4*23 


io-o45563 


60 


i 


82565i 


2 


.33 


870960 1 


.90 


934691 


4 


•23 


o453o9 


5o 


2 


823791 


2 


33 


870846 1 


■ 90 


934945 


4 


23 


045o55 


58 


3 


820931 


2 


33 


870732 1 


.90 


933200 


4 


■23 


044800 


57 


4 


826071 


2 


33 


870618 1 


.90 


935454 


4 


23 


044546 


56 


5 


82621 1 


2 


33 


870304 1 


90 


935707 


4 


•23 


044293 


55 


6 


82635i 


2 


33 


870390 1 


90 


955961 


4 


•23 


044039 


54 


7 


826491 


2 


33 


870276 1 


90 


9562i5 


4 


23 


043783 


53 


8 


826631 


2 


33 


870161 1 


90 


956469 


4 


23 


04353 1 


52 


9 


826-70 


2 


32 


870047 1 


9 1 


936723 


4 


23 


043277 


5i 


10 


826910 


2 


32 


869933 1 


9i 


956977 


4 


23 


o43o23 


5o 


ii 


9-827049 


2 


32 


9-869818 1 


9i 


9-957231 


4 


23 


10*042769 


49 


12 


827189 


2 


32 


809704 1 


9 1 


957485 


4 


23 


0425l3 


48 


'3 


827328 


2 


32 


869389 1 


9i 


937739 


4 


23 


042261 


47 


ii 


827467 


2 


32 


869474 1 


9i 


957993 


4 


•23 


042007 


46 


i5 


827606 


2 


32 


86 9 36o 1 


9i 


938246 


4 


•23 


041754 


45 


16 


827145 


2 


32 


869245 1 


9 1 


9585oo 


4 


23 


o4i5oo 


44 


n 


82^884 


2 


3i 


869130 1 


9' 


958754 


4 


23 


041246 


43 


18 


828023 


2 


3i 


8690 1 5 1 


92 


959008 


4 


•23 


040992 


42 


»9 


828162 


2 


3i 


868900 1 


92 


939262 


4 


23 


040738 


4i 


20 


8283oi 


2 


3i 


868785 1 


92 


959516 


4 


23 


040484 


40 


21 


9.828430 


2 


3i 


9-868670 1 


92 


9-959769 


4 


23 


io-o4o23i 


3 9 


22 


8285 7 8 


2 


3i 


868555 1 


92 


960023 


4 


23 


o39977 


38 


23 


828716 


2 


3i 


868440 1 


92 


960277 


4 


23 


039723 


37 


24 


828855 


2 


3o 


868324 1 


92 


96o53i 


4 


23 


039469 


36 


25 


828993 


2- 


3o 


868209 1 


92 


960784 


4 


23 


039216 


35 


26 


829131 


2' 


3o 


868o 9 3 1 


92 


961038 


4 


23 


038962 


34 


27 


829269 


2- 


3o 


867978 1 
867862 1 


9 3 


961291 


4 


23 


038709 


33 


28 


829407 


2' 


3o 


93 


96i545 


4 


•23 


o38455 


32 


29 


829045 


2 


3o 


867747 1 


93 


961799 


4 


23 


o3S2oi 


3i 


3o 


829683 


2 


3o 


867631 1 


9 3 


962052 


4 


23 


037948 


3o 


3i 


9.829821 


2- 


29 


9-86 7 5i5 1 


93 


9-962306 


4 


23 


10-037694 


20 


32 


829959 


2- 


29 


867399 1 


93 


962360 


4 


23 


037440 


28 


33 


830097 


2 


29 


867283 1 


93 


962813 


4 


•23 


037187 


27 


34 


830234 


2 


29 


867167 1 


93 


963067 


4 


23 


o36933 


26 


35 


83o372 


2- 


29 


867051 1 


93 


963320 


4 


23 


o3668o 


25 


36 


83o5o9 


2^ 


29 


866 9 35 1 


94 


963574 


4 


23 


o36426 


24 


37 


83o646 


2- 


29 


866819 1 


94 


963827 


4 


23 


o36i73 


23 


38 


830784 


2 


29 


866703 1 


94 


964081 


4 


■23 


035919 


22 


3 9 


830921 


2- 


28 


866586 1 


94 


964335 


4 


23 


o35665 


21 


40 


83 1 o58 


2" 


28 


866470 1 


94 


964588 


4 


22 


o354i2 


20 


41 


9 -83 1 195 


2- 


28 


9-866353 1 


94 


9-964842 


4 


22 


io-o35i58 


IS 


42 


83i332 


2- 


28 


866237 1 ■ 


94 


963095 


4 


22 


o349o5 


43 


83 1469 


2 


28 


866120 i- 


94 


963349 


4 


22 


o3465i 


17 


44 


83 1 606 


2- 


28 


866004 1 


9 5 


965602 


4 


22 


034398 


16 


45 


831742 


2 


2S 


863887 1 


9 5 


965855 


4 


22 


o34U5 


i5 


46 


831879 


2 


28 


865770 1 


9 5 


966105 


4 


22 


033S91 
033638 


14 


47 


832013 


2 


27 


865653 1 


9 5 


966362 


4 


22 


i3 


48 


832i 52 


2 


27 


865536 1 


9 5 


9666 1 6 


4 


22 


033384 


12 


49 


832288 


2 


27 


863419 i- 


93 


966869 


4 


22 


o3 3 1 3 1 


11 


5o 


832425 


2 


27 


8653o2 1 ■ 


93 


967123 


4 


22 


032877 


10 


5i 


9-83256i 


2 


27 


9 -865i85 i- 


9 5 


9.967376 


4 


22 


10-032624 


I 


52 


832697 


2 


27 


865o68 1 ■ 


9 5 


967629 


4 


22 


082371 


53 


832833 


2 


27 


864930 1 ■ 


93 


967883 


4 


22 


o3 2 1 1 7 


7 


54 


832969 


2 


26 


864833 1 


96 


968 1 36 


4 


22 


o3i864 


6 


55 


633 1 03 


2 


26 


864716 i- 


96 


9 6838 9 


4 


22 


o3 161 1 


5 


56 


833241 


2 


26 


864598 1 ■ 


96 


968643 


4 


22 


o3 1 357 


4 


12 


833377 


2 


26 


864481 I- 


96 


968896 


A- 


22 


o3 1 1 04 


3 


8335i2 


2 


26 


864363 1 ■ 


96 


969149 


4- 


22 


o3o85» 


2 


5 9 


833648 


2 


26 


864245 1 • 


96 


96940J 


4- 


22 


o3o597 


1 


60 


833783 


2-26 


864127 1 


96 


969636 


4-22 


o3o344 




M. 


Coeine 


D. 


Sine D. 


Cotsing. 

<- 


D. 


Tang. 



(47 DEGREES.) 



SINES AND TANGENTS. (43 DEGREES.; 



61 



M. 

o 


Sine 


D. 


Cosine 


D. 


Tang. 


'D. 


Cotang. 


60 


9-833783 


2-26 


9-864(27 


1-96 


9 969656 


4-22 


io-o3o344 


i 


833919 


2-25 


864010 


1-96 


9°99°9 


4-22 


o3oo9( 


bo 


2 


834034 


2-25 


8638 9 2 


1-97 


970162 


4-22 


02 9 838 


58 


3 


834i8o 


2-25 


863 77 4 


1.97 


9704(6 


4-22 


029584 


57 


4 


83432D 


2-25 


863656 


1-97 


970669 


4'22 


02933 ( 


56 


5 


83446o 


2-25 


863538 


1-97 


970922 


4-2* 


029078 


55 


6 


834295 


2-25 


863419 


1.97 


97(175 


4-22 


028825 


54 


I 


83473o 


2-25 


8633o( 


1-97 


97U29 


4-22 


02857( 


53 


834865 


2-25 


863(83 


1.97 


97(682 


4-22 


028318 


52 


9 


834999 


2-24 


863o64 


1-97 


97(935 


4-22 


028060 


5i 


10 


835i34 


2-24 


862946 


1-98 


972(88 


4-22 


0278(2 


5o 


ii 


9-835269 


2-24 


9-862827 


1-98 


9-972441 


4-22 


10-027559 


49 


12 


8354o3 


2-24 


862709 


1.98 


972694 


4-22 


027306 


48 


i3 


835538 


2-24 


862590 


1-98 


972948 


4-22 


027052 


47 


14 


835672 


2-24 


862471 


1.98 


973201 


4-22 


026799 


46 


i5 


8358o 7 


2-24 


862353 


1-98 


973454 


4-22 


026546 


45 


16 


835941 


2-24 


862234 


1-98 


973707 


4-22 


026293 


44 


\l 


836075 


2-23 


862(i5 


1-98 


973960 


4-22 


026040 


43 


836209 


2-23 


861996 


1-98 


9742(3 


4-22 


025787 


42 


19 


• 836343 


2-23 


86(877 


1-98 


974466 


4-22 


025534 


4i 


20 


836477 


2-23 


86(758 


1.99 


974719 


4-22 


025281 


40 


21 


9-8366H 


2-23 


9-86(638 


1.99 


9-974973 


4-22 


10-025027 


3 9 


22 


836745 


2-23 


86(5(9 


1-99 


975226 


4-22 


024774 


38 


23 


8368 7 8 


2-23 


86(400 


1-99 


975479 


4-22 


02452( 


37 


24 


837012 


2-22 


86(280 


1-99 


975732 


4-22 


024268 


36 


25 


83 7 i46 


2-22 


86((6i 


1-99 


975985 


4-22 


0240(5 


35 


26 


837279 


2-22 


86(041 


1-99 


976238 


4-22 


023762 


34 


2 I 


837412 


2-22 


860022 


1-99 


976491 


4-22 


023509 


33 


28 


837546 


2-22 


860802 


1-99 


976744 


4-22 


023256 


32 


29 


837679 


2-22 


860682 


2-00 


976997 


4-22 


o23oo3 


3i 


3o 


837812 


2-22 


86o562 


2-00 


977250 


4-22 


022750 


3o 


3i 


9-837945 


2-22 


9-860442 


2-00 


9-9775o3 


4-22 


10-022497 


29 


32 


838078 


2-21 


86o322 


2-00 


977756 


4-22 


022244 


28 


33 


83821 1 


2-21 


860202 


2- OO 


978009 


4-22 


02(991 


?7 


34 


838344 


2-21 


860082 


2-00 


978262 


4-22 


02(738 


26 


35 


838477 


2-21 


859962 


2-00 


9785(5 


4-22 


021485 


25 


36 


8386io 


2-21 


85g842 


2-00 


978768 


4-22 


02(232 


24 


ll 


838742 


2-21 


859721 


2-01 


979021 


4-22 


020979 


23 


838875 


2-21 


859601 


2-01 


979274 


4-22 


020726 


22 


3 9 


839007 


2-21 


85 9 48o 


2-01 


979527 


4-22 


020473 


2( 


40 


839140 


2-20 


85 9 36o 


2-01 


979780 


4-22 


020220 


20 


4i 


9-839272 


2-20 


9-859239 


2-01 


9-980033 


4-22 


10-0(9967 


IO 


42 


839404 


2-20 


8591 [9 


2-01 


980286 


4-22 


0197(4 


l8 


43 


839536 


2-20 


858 99 8 


2-01 


9 8o538 


4-22 


0(9462 


\l 


44 


819668 


2-20 


8588 77 


2-0( 


980791 


4.21 


0(9209 


45 


839800 


2-20 


858 7 56 


2-02 


981044 


4-21 


0(8956 


(5 


46 


839932 


2-20 


858635 


2-02 


98(297 


4-2( 


0(8703 


i4 


4 2 


840064 


2- (9 


8585(4 


2-02 


981550 


4-21 


Ol845o 


i3 


48 


840196 


2- 19 


858393 


2-02 


981803 


4-21 


018(97 


12 


i 9 


840328 


a. 19 


858272 


202 


982056 


4-2( 


017944 


:( 


5o 


840459 


2-19 


858(5i 


2-02 


982300. 


4-21 


0(7691 


10 


5i 


9-840591 


2-19 


9.858029 
857908 


2-02 


9-982502 


4-2( 


10-017438 





52 


840722 


2-19 


2-02 


9828(4 


4-21 


oi4i86 


i 


53 


84o854 


2-19 


857786 


2-02 


983067 


4-21 


016933 


7 


54 


840985 


2-IO 


857665 


2-o3 


9 8332o 


4-21 


1 6680 


6 


55 


841 1 16 


2-l8 


857543 


2-o3 


983573 


4-2( 


016427 


5 


56 


841247 


2-(8 


857422 


2 03 


983826 


4-21 


016174 


4 


& 


84i3 7 8 


2-l8 


8573oo 


2-o3 


984079 


4.21 


01 5921 


3 


841509 


2-(8 


85 7 ( 7 8 


2-o3 


9843! 1 


4-21 


1 5669 


2 


59 


84 1 640 


2-(8 


857056 


2-o3 


9'' 


4-2( 


01 5/, 16 


1 


60 


84177' 


2-(8 


856934 


2-o3 


984837 


4-2( 


oi5i63 




M. 


Cosine 


D. 


Sine 


1). 


Cotani; 


1). 


Tang. 



(46 DEGREES.) 



62 



(44 DEGREES.) A TABLE OF LOGARITHMIC 



M. 



o 
i 

2 

3 

4 
5 
6 

7 
8 

9 

10 

ii 

12 

i3 
14 
i5 
16 

17 
18 

19 

20 

21 
22 
23 

24 
25 
26 



Sine 



D. 



Cosine 



8 

3i 

32 

33 
34 
35 
36 

\l 

38 

3 9 
4o 

4i 
42 
43 
44 
45 
46 

47 
48 

49 
5o 

5i 

52 

53 
54 
55 
56 

a 

59 
6o 



9-84I77 1 
841902 
842033 
842i63 
842294 
842424 
842555 
842685 
8428i5 
842946 
843076 

9.843206 
843336 
843466 
843595 
843725 
843855 
843984 
844i 14 
844243 
844372 

g. 844502 

844631 

8447 6 o 
844889 
845oi8 

845i47 
845276 
8454o5 
845533 
845662 

9.845790 
845919 
8460/7 
846n5 
8463o4 
846432 
84656o 
846688 
8468 1 6 
846944 

9.847071 

847199 
847327 

847454 
847582 

8477 9 
847836 

847964 
848091 
848218 

•848345 
848472 
848599 
848726 
848852 

848979 
849106 

849232 

8493 5g 

84948a 



2-i8 
18 
18 

17 
17 
17 
17 

17 
'7 
«7 

16 
.16 
.16 
.16 
.16 
.16 
.16 

• i5 

• i5 

• i5 



D. 



2-13 
2-l5 
2-l5 
2-l5 
2«l5 
2-l5 
2-14 
2-14 
2-14 
2-14 



2- 

2- 

2- 

2- 

2- 

2- 

2' 

2 

2 

2 

2 
2 
2 
2 

2 

2 

2- 

2- 

2- 

2- 

1- 

2' 

2 

2 

2 

2 

2 

2 

2 

2 



U 

14 

14 
14 
U 

i3 
i3 
i3 

• i3 
i3 

• i3 

• i3 

• i3 

• 12 
■ 12 

• 12 
12 
12 
12 
12 

12 
11 
11 

• 11 

• 11 

• 11 

• 11 

• 11 

• 11 

• 11 



,.856934 
8568i2 
836690 
856568 
856446 
856323 
856201 
856078 
855956 
855833 
85571 1 

9-855588 
855465 
855342 
855219 
855096 

854973 
85485o 

854727 
8546o3 
85448o 

9-854356 
854233 
854109 
853986 
853862 
853738 
8536U 
853490 
853366 
853242 

9 -853ii8 

852994 
852b6 ? 
852743 

852&20 

852496 
852371 
852247 

852122 

85 1997 

9.851872 

85 1 747 
85i622 

85 1 497 
85i372 

85 1 246 

85 r 1 2 1 

850996 
850870 
85o745 

i-85o6i9 
85o493 
85o368 
85o242 
85oii6 
849990 
849864 
849738 
8496 1 1 
849485 



2-o3 
2-o3 

2- 04 



Tang. 



2 

2- 

2- 

2' 

2 

2 

2 

2 

2 
2 
2 
2 
2 
2 
2 
2 
2 



04 

04 

>04 

.04 
.04 
•04 
.04 
• o5 

•o5 

•o5 

•o5 

•o5 

•o5 

•o5 

o5 

06 

06 



2- 06 

2- 06 
2- 06 
2- 06 
2- 06 
2- 06 
2- 06 
2-07 
2-07 
2-07 
2-0-J 



2- 


07 


2- 


07 


2- 


07 


2- 


07 


2 


07 


2 


08 


2 


08 


2 


08 


2 


•08 


2 


•08 


2 


•08 


2 


-08 


2 


•08 


2 


•09 


2 


•09 


2 


•09 


2 


•09 


2 


•09 


2 


•09 


2 


•09 



D. 



09 
10 
10 
10 
10 
10 
10 
10 
10 
10 



9-984837 
085090 
985343 
985596 
985848 
986101 
986354 
986607 
986860 
9871 12 
987365 

9.987618 

987871 
988123 
988376 
9S8629 
988882 
989134 
989387 
989640 
989893 

9-990143 
990398 
990631 
990903 
991 1 56 
991409 
991662 

9919U 
992167 
992420 

^•992672 
992925 
993178 
99343o 
9 9 3683 
993936 
994189 
994441 
994694 
994947 

9-995199 
995432 
995705 
993937 
996210 
996463 
9967 1 5 
996968 
997221 
997473 

9.997726 

997979 
998231 

998484 
998737 
998989 
999242 
999495 
999748 

10-000000 



4- 
4- 
4- 
4- 
4- 
4' 
4' 
4' 
4 
4 
4 



21 

21 
21 
21 
21 
21 
21 
21 
21 
21 
21 



4-21 

4-21 



4- 

4- 

4- 

4' 

4' 

4 

4 

4 

4- 
4- 
4- 
4- 
4- 
4- 
4- 
4- 
A- 
4- 

4- 
4' 
4' 
4' 
4 
4 
4 
4 
4 
4 



21 
21 
21 
21 
21 
21 
21 
21 

21 

21 
21 
21 

■21 
•21 
•21 
•21 
21 
21 

21 
21 
21 
21 
21 
21 
21 
21 
21 
21 




4 

4 

4 

4 

A- 

A; 

A- 

4- 

4- 

4- 

4- 

4- 

4- 

4- 

4' 

4' 

4' 

4 

4 

4 



Cotang. 



21 

21 
21 
21 
21 
21 
21 
21 
21 
21 

21 
21 
21 

■ 21 

• 21 

■ 21 

• 21 

• 21 
21 
21 



1 • 1 5 1 63 
1 49 1 
014657 
014404 
0i4i52 
013899 
oi3646 
oi3393 
o 1 3 1 40 
012888 
012635 

10-012382 

012129 
01 1877 
01 1624 
01 1 37 1 
01 1 1 18 
1 0866 
01 061 3 
oio36o 
010107 

10-009855 
009602 
009349 
009097 
008844 
008591 
oo8338 
008086 
007833 
007580 

10-007328 

007073 
006822 
006570 
006317 
006064 
oo58n 
oo5559 
oo53o6 
oo5o53 

10-004801 

004348 
004295 
004043 
003790 
oo353] 
oo3285 
oo3o32 
002779 
002527 

10-002274 
002021 
001769 
001 5 16 
001263 
00101 1 
000758 
ooo5o5 

000233 
10-000000 



60 
5o 
58 

57 
56 

55 
54 
53 

52 

5i 
5o 

49 
48 

47 
46 
45 
44 
43 
42 
41 
4o 

3 

n 

35 
34 
33 

32 

3i 
3o 



2C 

2< 

27 

26 

25 

24 
23 
22 
21 
20 

IO 
10 

17 
16 

i5 

14 
i3 
12 
11 
10 



I 

5 
4 
3 
2 
I 
o 



(45 DEGREES.) 



